t o •a I. I.Gcl dman V. D. Krivchenkov V. I. Kogan V.M.Galitskii PROBLEMS IN QUANTUM MECHANICS Translated, edited and arranged by D. ter Haar infosearch Infosearch Pwtt'cs £>£Pr. 1. 1. Gol 'dman V. D. Krivchenkov V.I.Kogan V. M.Galitskii PROBLEMS IN QUANTUM MECHANICS Translated, edited and arranged by D* ter Haar d> Infosearch Limited London Translated from Russian 1 1. GOL' OMAN and V. D. KR1VCHENKO V Sbornik Zadach po Kvantovoi Mekhanike supplemented by a selection from V.I.KOGAN and V. M. GALITSKII Shoraik Zadach po Kvautovoi Mekhanike Gosudarstvennoe IidateL'atvo Tekhniko-Teoretieheskoi Literatury 276 + 416 pp., Moscow, 1956 Copyright I960 by Iafosearch Ltd. ALL EIGHTS RESERVED NO PART OP THIS BOOK MAY BE REPRODUCED IN ANY FORM BY PHOTOSTAT, MICROFILM, OR ANY OTHER MEANS, WITHOUT WRITTEN PERMISSION PROKt THE PUBLISHERS INFOSEARCH LTD. 242 Willesden Lane London, N.W.2. Distributed outside the Western Hemisphere by CLEAVER-HUME PRESS LTD 31 WRIGHT'S LANE, LONDON w8 Printed in Great Britain by Pion Ltd. CONTENTS TRANSLATOR'S PREFACE 1 AUTHORS' PREFACE 3 PROBLEMS 1. One -Dimensional Motion; Energy Spectrum and Wave Functions 7 2. Tunnel Effect 11 3. Commutation Relations; Heisenberg Relations; Spreading of Wave Packets; Operators 15 4. Angular Momentum; Spin 22 5. Central Field of Force 29 6. Motion of Particles in a Magnetic Field 32 7. Atoms 35 8. Molecules 45 9. Scattering 49 ANSWERS AND SOLUTIONS 1. One-Dimensional Motion; Energy Spectrum and Wave Functions 59 2. Tunnel Effect 96 3. Commutation Relations; Heisenberg Relations; Spreading of Wave Packets; Operators 123 4. Angular Momentum; Spin 179 5. Central Field of Force 194 6. Motion of Particles in a Magnetic Field 212 7. Atoms 225 8. Molecules 302 9. Scattering 331 APPENDIX I 383 APPENDIX n 387 INDEX 391 TRANSLATOR'S PREFACE This collection contains all problems from the text by Gol'dman and Krivchenkov, as well as a selection from that by Kogan and Galitskii, The latter are arranged in the corresponding chapters of the Gol'dman and Krivchenkov collection and distinguished by asterisks, Although these problems are used in the U.S.S.R. with Landau and Lif shits' "Quantum Mechanics", they can be used equally well with any other textbook on quantum mechanics, such as those by Schiff or Kramers, or as advanced reading by students familiar with the basic ideas of quantum mechanics, e.g. acquired from Pauling and Wilson's text. In order to keep the price of this book down to a minimum, all for- mulae have been reproduced photographically from the Russian originals. There is thus a slight difference in notation between Gol'dman and Krivchenkov 'a and Kogan and Galitskii* s collections. Moreover, there are a few symbols different from those commonly used in English texts. We give here a list of the more important ones: The vector product of two vectors a and b is indicated in this volume by: [a b], ax b 7 or [a x b) The dot or scalar product of two vectors a and b is indicated in this volume by ab or (ab) The Laplacian V 2 is indicated here by A. The anticommutator [A, &]_ {s Afc - &&) is indicated by [A, fe] or {A, §} All operators are distinguished by a circumflex -. Finally, tg, ctg, ch, sh, th, denote respectively tan, cot, cosh, sinh, and tanh. DAe.rH.aar AUTHORS' PREFACE This collection consists of problems in non-relativistic quantum mechanics of varying degrees of difficulty, which were solved in seminars or were given as exercises to fourth -year students of the physical faculty of the Moscow State University. The more elaborate problems were intended mainly for students specialising in theoretical physics whose course was based on L.D. Landau and E. M. Lifshits* textbook "Quantum Mechanics". Teaching experience shows that greatest difficulties are encountered in the matrix aspect of quantum mechanics so that in composing this collection much attention has been devoted to constructing the perturba- tion matrix and diagonalising it. Relatively much space has also been accorded to angular momentum and spin since without an understanding of these fundamental ideas it is impossible to speak of a serious study of quantum mechanics, l.l.GoVdman V.D.Krivckenkov PROBLEMS PROBLEMS * 1. ONE -DIMENSIONAL MOTION. ENERGY SPECTRUM AND WAVE FUNCTIONS 1. Determine the energy levels and the normalised wave functions of a particle in a "potential well". The potential energy V of the particle is: K = oo forx<0 and for x>a, K = for < x < a. 2. Show that for particles in a "potential well" (see preceding problem) the following relations hold: 5-4 a. ( X --xY = £(l-^). Show also that for large values of n the last result agrees with the corresponding classical result. 3. Determine the momentum probability distribution function for particles in the n-th energy state in a "potential well". J Vrxi 4. Determine the energy levels and wave functions of a particle in an 1 V t asymmetrical potential well (see fig. 1). Consider the case where V l = V 2 . 5. The Hamiltonian of an oscilla- tor is equal to // = 2jc2 , where p d Fig.l 2^ ' 2 and x satisfy the commutation rela- tionships px — xp = — ih. In order to get rid of h, jx,. and co in the following calculations, we introduce new variables P and Q ** P = fifiu Q V [A(0 X (PQ—QP == — i ) t * Problems which are starred are taken from a similar collection by V.I. Kogan and V. M. Galitskii, 1956. ** All operators are indicated by a circumflex " . 8 PROBLEMS and the energy E will be expressed in units tm (E = efiu>). The SchrOdinger equation for the oscillator in the new variables will be of the form /fy = _(p2 + Q2 )(1>=ef a) Use the commutation relation PQ—QP=—i, to show that Y ( p2 + Q 2 ) (Q =t iPf*? =(e + n)(Qz±: iP) n <|>. b) Determine the normalised wave functions and the energy levels of the oscillator. c) Determine the commutator of the operator a =-rL-(Q+/P) i » and its Hermitean conjugate operator a+ = -— (Q — iP). Express the wave function of the h-th excited state in terms of the wave func- tion of the ground state using the operator a. d) Determine the matrix elements of the operators P and Q in the energy representation. Hint . P 2 + Q 2 — 1 = (P + IQ) (P — tQ). 6. Using the results of the preceding question, show by direct multiplication of matrices that for an oscillator in the n-th stationary state we have 7. A particle moves in a potential V(x) = ^^_. Etetermine the probability to find the particle outside the classical limits, when it is in its ground state. 8. Find the energy levels of a particle moving in a potential of the following form : V(*) = oo (*<0); V(x) = -^ (*>0). 9. Write down the SchrOdinger equation for an oscillator in the "^-representation" and determine the momentum probability distri- bution function. 10. Find the wave functions and energy levels of a particle in a potential V(x) = V Q (~ — ~) 2 (jc > 0) (see fig. 2) and show that the energy spectrum is the same as the oscillator spectrum. 11. Determine the energy levels for a particle in a potential y — K ° (see fig. 3). cha^ a ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 12. Determine the energy levels and wave functions for a par- ticle in the potential V = V Q ctg 2 ^-x (0<x<a) (see fig. 4), and derive the normalisation constant of the ground state wave function. Consider the limiting cases of small and large values of V . 13. Determine the wave fun- ctions of a charged particle in a uniform field V(x) = — Fx. 14. Write down the SchrSdin- ger equation in the ' '^-represen- tation u for a particle moving in a periodic potential V (x) = V cos bx. 1 5 . Write down the Schr 5din - ger equation in the "/?-represen- tation" for a particle moving in a periodic potential V(x) = V(x-\-b). 16. Determine the allowed energy bands of a particle moving in the periodic potential given by fig. 5. Investigate the limiting case where V -+ oo, and b -> while V b = const. Fig. 4 17. Determine the energy levels of the potential V = Vo cifi — a in the semi-classical approximation, and also the total number of discrete levels. 18. Determine in the semi -classical approximation the energy spectrum of a particle in the following potentials: a) V (J.U>"JC 3 ,2 (oscillator); b) V=V ctg^x (0<x<a). 10 PROBLEMS 19. Determine in the semi-classical approximation the average value of the kinetic energy in a stationary state. 20 . Use the result of the preceding question to find in the semi -classical approxi- mation the average kinetic energy of a particle in the following potentials: i V(xi i n_ -b a a+b x a) y. Fi g- 5 b) V = V ctg 2 — (0 < x < a) (see question 18). 21. Determine the form of the energy spectrum of a particle in a potential V(x) = ax' t , using the semi-classical approximation and applying the virial theorem. 22. Determine in the semi -classical approximation the form of the potential energy V(x) for a given energy spectrum E n . V(x)may be assumed to be an even function V(x) = V( — x), which increases monotonically for x > 0. 23. * Find the wave functions and energy levels of the stationary states of a plane rotator with moment of inertia /. A rotator is a system of two rigidly connected particles rotating in a plane (or in space). The moment of inertia of a rotator is equal to / = M-a a , where ^ is the reduced mass of the particles and a their distance apart. 24.* Find the wave functions and energy levels of the stationary states of a particle of mass m in a uniform gravitational field g for the case when the region of the motion of the particle is limited from below by a perfectly reflecting plane. (As a classical analogy of this system we can take a heavy solid ball, bobbing up and down on a metallic plate. We note that all calculations and results of this pro- blem are clearly correct also for the case of the motion of a particle of charge e in a uniform electric field & , in the presence of a re- flecting plane, provided we replace in all equations g by — t. ) Take the limit to classical mechanics. 25.* A particle is enclosed in a one -dimensional rectangular potential well with infinitely high walls. Evaluate the average force exerted by the particle on the wall of the well. 26.* A particle in an infinitely deep rectangular potential well is in a state described by the wave function ty(x) = Ax(a — x), where a is the well depth and A a constant. TUNNEL EFFECT H Find the probability distribution for the different energies of the particle and also the average value and the dispersion of the energy. 27.* Obtain the semi -classical expression for the energy levels of a particle in a uniform gravitational field for the case where its motion is limited from below by a perfectly reflecting plane. 28.* A particle moves in a periodic field U(x): U(x-\-a) = U(x). Using a suitable semi-classical approximation obtain a tran- scendental equation to determine the allowed energy bands. Discuss this equation. 29.* Find the semi -classical solution of the Schrodinger equa- tion in the momentum -representation. Show that the same semi -classical function is obtained by going over from the "at -representation" to the "jo-representation" starting from the usual semi-classical coordinate wave function. 2. TUNNEL EFFECT 1. In studying the emission of electrons from metals, it is necessary to take into account the fact that electrons with an energy sufficient to leave the metal may be reflected at the metal surface. Consider a one -dimensional model with a potential V which is equal to — V for x > (inside the metal) and equal to zero for x > (outside the metal) (see fig. 6), and determine the reflexion coefficient at the metal surface for an electron with energy E > . 2. In the preceding question it was assumed that the potential changed discontinuously at the metal surface. In rea- lity this change in potential takes place continuously over a region of the dimensions of the order of the interatomic distance in the metal. Approximate the potential near the metal surface by the function y = ^— (see fig. 7) e a +1 I War; x Fig. 6 12 PROBLEMS and determine the reflexion coefficient of an electron with energy £>0 3. Determine the coefficient of transmission of a particle through a rectangular barrier (see fig. 8). 4. Determine the coeffi- cient of reflexion of a particle by a rectangular barrier in the case where E > v (reflexion above the barrier) 5. Fig. 7 Calculate the coefficient of transmission through a potential barrier V(x)= -J^. ( see fig. 9) for a flux of particles moving with an energy E < v ( Y( X j m Fig. 8 " Fig. 9 6. Calculate in the semi -classical approximation the coeffi- cient of transmission of electrons through a metal surface under the action of a large electrical field strength F (fig. 10). Find the limits of applicability of the calculation. 7. The change of the potential near a metal surface is in reali- ty a continuous one. For instance, the electrical image potential K.im. = — ^ will act at large distances from the surface. Deter- mine the coefficient of transmission D of electrons through a metal surface under the action of an electrical field, taking into account the electrical image force (fig. II) 8. Determine approxi- mately the energy levels and wave functions of a particle in the symmetrical potential given by fig. i2 for the case where Fig. io TUNNEL EFFECT 13 £<C Vq and the penetrability of the barrier is small ( 2 ^ v o ^ -^ A •^ 9. A symmetrical potential V(x) consists of two potential wells separated by a barrier (see fig. 13). Assuming that one may use a semi- classical argument, determine the energy levels of a particle in the potential V(x), Com- pare the energy spectrum obtained with the energy Fig. 11 spectrum of a separate well, of a separate well. Hint. See Appendix I. Find the splitting of the energy levels V(X) -E a+b Za+b & Fig. 12 , J(XJ 1 1 1 1 1 i , 1 1 n^ S L -b -a 0+a *b X Fig. 13 10. Assume that at t = there exists an impenetrable partition between the two symmetrical potential wells (see preceding question) and that a particle is in a stationary state in the well on the left. Determine after how long a time t after the partition is removed the particle will be in the well on the right. 11. The potential V(x) consists of N identical potential wells separated by identical potential barriers (see fig. 14). Determine the energy levels in this potential, assuming that one can use the semi-classical approach. Compare the energy spectrum obtained with the energy spectrum of a separate well. 14 PROBLEMS , , VlX) 12. Assuming that one may use the semi -classical approach, find the quasi-stationary levels of a particle in the symmetrical field given by fig. 15. Find also the transmission co- efficient D(E) for a particle with energy E < v , where V is the maximum value of the potential V{x). 13.* Show generally that for Fig. 14 any barrier the relation /?-f-D=l, is automatically satisfied, where R is the reflexion coefficient, and D the transmission coefficient. 14.* Find the coefficient of- trans- mission of a particle through a triangular barrier (see fig. 16). Consider the limiting cases of small and of large penetrability. Fi 8- 15 15.* Evaluate in the semi -classical approximation the trans- mission coefficient for a parabolic potential barrier of the follow- ing form (see fig. 17): , ' t A r* -b -<Vj k b » £/(*) for for \x\>a. (1) Give the criterion for the applicability of the result obtained. COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 15 3. COMMUTATION RELATIONS; HEISENBERG RELATIONS; SPREADING OF WAVE PACKETS; OPERATORS 1. Show that if the two Hermitean operators A and B satisfy the commutation relation AB — BA = iC, the following relation will hold K(Ai)2(Afi)2>i|i. 2. Find the uncertainty relation for the operators q and F(p), if q and p satisfy the commutation relation qp — pq = ih. Hint. Express F(p) in a Taylor series. 3. Use the Heisenberg relations to estimate the ground state energy of a harmon.c oscillator. 4. Estimate the energy of an electron in the tf-shell of an atom of atomic number Z both for the relativistic and the non-relativistic case. 5. Estimate the ground state energy of a two-electron atom with nuclear charge Z, using the Heisenberg relations. 6. The magnetic field produced by a free electron is partly due to its motion, and partly due to the presence of its intrinsic mag- netic moment. It is known from electrodynamics that the magnetic field strength due to a moving charge is of the order of magnitude and the magnetic field strength due to a dipole moment jt H r To determine the magnetic moment n of a free electron from a measurement of the field strength produced by it, it is necessary that the following two conditions are satisfied : 16 PROBLEMS and Ar<Cr. (2) The meaning of the last condition is that the electron must be loca- lised in a region Ar which is much smaller than the distance from that region to the point where the magnetic field is observed. Is it possible to satisfy these two conditions simultaneously? Hint. Take into account the Heisenberg relations and the value all of the electronic magnetic moment u. = 2~"c* 7. What is the physical meaning of the quantity p in the expression for the wave function if <p(jc) is a real function? 8. Show that the average value of the momentum in a stationary state with a discrete energy eigenvalue is equal to zero. 9. The wave function of a free particle at t = is given as follows iv%x ^( X , 0) = <f(x)e~ ir . The function <p (x, 0) is real and appreciably different from zero only for values of x, within the interval — 8 < x < 4-8. Determine in which region of x -values the wave function will be different from zero at time t. 10. Find the change of a wave function which is given at t = (spreading of a wave packet) for the following three cases: a) a free particle TV («W)'/. F \ h 282 /' b) a particle moving in a uniform field TV (*W)'/« H \ ft 262/' c) a particle moving in a potential V = ** .* - 11. Prove the following relation e £^-£ = aH-[La]-h^[^|/^l]-h^[^[^lM]]+ ... COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 1*7 12. An oscillator is at t — — oo in its ground state. Determine the probability that at /== + oo the oscillator is in the »-th excited state, if it is acted upon by a force /(/), where /(*) is an arbitrary function of the time (f=0 fo r /-»-=t:cx>). Evaluate the expression obtained for a > /(0 = / «" t ~, b) /(0=/o 7T^- 13. Show that the problem of how to determine the motion of an oscillator under the action of an external force /(/) can be reduced to the very simple problem of determining the motion of a free oscil- lator, if we introduce the new variable x t = x~ £ (/), where £(0 satisfies the classical equation ji£=/(*) — {WS. 14. Find the Green function for an oscillator whose eigen frequency changes with time, and express it in terms of the solution of the classical equation of motion of an oscillator of varying frequency. 15. Use the Green function obtained in the preceding question to determine the change in time of the probability density for par- ticles moving in a potential V{x) = ^- («, = const). The particle wave function at / = is equal to ty(x, 0) = ce 2 » . 16. Find the Green function of an oscillator whose eigen frequency changes in time and which is acted upon by a perturbing force f(t). 17. An oscillator is at t = in its n-th energy eigen state. Determine the probability that it will make a transition to the m -th eigen state under the action of a perturbing force /(*). Find the average value and the dispersion of its energy at time t. 18. Inasmuch as the SchrOdinger equation is a first order dif- ferential equation with respect to time, <K0 is uniquely determined by the value of <K0). Write this connection in the form *(0 = ^(0*(0), where S(t) is some operator. a) Show that the operator S(t) satisfies the equation ihS(t) = HS(t) and is a unitary operator, that is, S+ = S~ 1 . Ifc PROBLEMS b) Show that in the case where H does not depend on time, S(t) is of the form i6t i9. The average value of an operator L at time t follows from the following expression Z(*)=JV(0£i>(0<ft- a) Show that the time dependence of the operator jp = S- i (t)LS(t), with S(t) determined bv the equation 5(0<!»(0) = <H0> satisfies the equation J«V(0)^(0)dx = L(0- b) Prove the following operator equation: ihJ'^J'SW — SVj', where S& = S~ l HS. c) Show that if the operators I and M satisfy the commutation relation LM-ML^iN, the corresponding time dependent operators satisfy the equation 20. Determine the time dependence of the coordinate operator x (in the coordinate representation) for (a) a free particle, and (b) an oscillator. 21. Use the result of the preceding question to determine the time dependence of the dispersion of the coordinate for the case of a free particle. 22. The wave function of a particle at / = is of the form ^ ( X ) _ jp ( X ) e k t w here cp(x) is real and normalised. Determine the value of the dispersion "(Ax) 2 at time / for the cases (a) and (b) of question 20, section 3. Show that in the case of the oscilla- tor "(Ax)* =(A;t) 2 =0 , that is, the dispersion does not increase, if y^ — ce tT (see question iOc, section 3). 23.* Let A, B and C be three operators. Express the com- mutator of the product AB with C in terms of the commutators [X C] and IB, C\. 24.* Show that for algebraic manipulations with commutators the distributive law holds, that is, that the commutator of a sum is COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 19 the sum of commutators: L * * J <. * 25.* Derive a law for the differentiation with respect to the time of the product of two operators. 26.* Show that the average value of the square of a Hermitean operator is positive. 27.* Does taking the complex conjugate correspond to: 1) a linear operator, 2) a Hermitean operator, 3) an operator which is its own complex conjugate? 28.* Assume A, to be a small quantity to find an expansion of the operator (^ — XB)' 1 i n powers of X. 29.* In a state of a quantum mechanical system described by a given wave function W A , the mechanical quantity A has a well-defined value. Can the quantity B also have a well-defined value in the case where the operators A and B 1) do not commute, 2) commute? 30.* Show that if the Hamiltonian of a system is invariant under some coordinate transformation, the operator of this transformation will commute with the Hamiltonian. 31.* Indicate which mechanical quantities or which combination of them are conserved when a system of N particles moves in the following external fields (consider energy, total angular momentum and its components, the components of the linear momentum, and parity): i) free motion; 2) field of an infinite, homogeneous cylinder; 3) field of an infinite, homogeneous plane; 4) field of a homogeneous sphere; 5) field of an infinite, homogeneous half -plane; 6) field of two points; 7) a uniform, variable field; 8) field of a conductor with variable charge; 9) field of a tri -axial ellipsoid; 10) field of an infinite, homogeneous screw line; 11) field of an infinite, homogeneous prism; 12) field of a homogeneous cone; 13) field of a cylindrical torus. Ascertain also the commutability of the corresponding operators. 32.* We consider a particle in a one -dimensional symmetrical potential well in which there is always, as is well-known, at least 20 PROBLEMS one energy level. If for a given depth of the well U its width a is reduced until it satisfies the inequality ht then, at first sight, the spatial localisation of a particle bound in the well will become much more precise (Ax — a), and as the spread in momentum A/> in any case is limited to a value of the order VmW , the following inequality tipLx^VW Q -a<g.h, would hold, violating the Heisenberg relations. Show the error in the above argument and evaluate the product of the spread in coordinate and the spread in momentum of the particle. 33.* Express the operator i. in the "/^-representation" and 1 r the operator — in the " r-representation". 34.* Express the operator — in the "^-representation" and - Px the operator — in the "/>„, -representation". 35.* Find the matrices of the coordinate and the momentum in the energy representation for a particle in an infinite one- dimensional rectangular potential well. 36.* Show that the average value of the time derivative of a physical quantity, which does not explicitly depend on the time, is equal to zero in a stationary state of the discrete spectrum. 37.* Prove the virial theorem in quantum mechanics. 38.* What is the change in time of the state of a plane rotator if at /«= it is described by a wave function W = A sin y <p ? . 39.* A particle is bound in an infinite rectangular potential well of width a. At / = one of the walls of the well starts to move according to an arbitrarily given time dependence. Reduce this problem to a wave equation with a Hamiltonian which depends explicitly on the time and discuss the particular case of the motion of the wall for which the variables in that equation can be separated. 40.* Prove the invariance of the non-relativistic Schrodinger equation under a Galilean transformation. 41.* Find the coordinate operator for the case of a free par- ticle in the Heisenberg representation. 42.* Find the coordinate and momentum operators in the Hei- senberg representation for the case of a linear harmonic oscillator COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 21 by solving the equations of motion for these operators. 43.* Prove that the function *, which at * = describes a state with a well-defined value of the physical quantity A, will at t > be an eigenf unction of the operator A( — t) corresponding to the same eigenvalue. The quantity A(y) is the operator corresponding to the quantity 4 in the Heisenberg representation at time t. 44.* The point of suspension x of an oscillator in its ground state starts to move at t = Q. The point of suspension moves accor- ding to the law x — x (f). At /= T the point of suspension is fixed again. Use the Green function (see problem 14, section 3) to find the wave function of the system at any time t > 0, and also the proba- bility of excitation of the n-th level as a result of the process con- sidered. Consider the limiting cases of fast and slow processes (o»T <C 1 and o>r^> 1, where oi is the frequency of the oscillator). 45.* The Hamiltonian H of a system performing a one- dimensional finite motion depends explicitly on the time. At every time t we assume that we know the eigenvalues E n {t) of the "instantaneous" Hamiltonian and the corresponding complete ortho- normal set of eigenfunctions $„(*)• (The term "instantaneous" is used here in the sense of "at a given time" which is different from its use in problems 74 and 75, section 7, which it is synonymous with "sudden". The dependence of the ^-function on the coordinate (q) is omitted here for the sake of simplicity. ) Give the wave function of the system in the representation whose basis is the set of functions <}»„(/). 46.* Let the Hamiltonian of the system, characterised in the preceding problem, be a slowly varying function of the time /. Assume that the system is at t = in the m -th quantum state and find its wave function at /> in the first approximation of the adia- batic perturbation theory and discuss the conditions for the appli- cability of the result obtained. 47.* The point of suspension of a linear oscillator in its ground state starts to move slowly at t = and comes to rest again at t=T. Find the probability for the excitation of the oscillator in the adiabatic approximation (compare problem 44, section 3) and ascertain the applicability of this approximation. 48.* A particle is in the n-th stationary state of an infinite rectangular potential well of width a. At * = one of the walls of 22 PROBLEMS the well starts to move slowly according to a given time dependence. Find in the adiabatic approximation the probability that at t > the particle will be found in the m-th stationary state (m ¥=n). 4. ANGULAR MOMENTUM; SPIN 1. Obtain expressions for the operators l x , l v , l t in spherical coordinates starting from the fact that i x , l y , l g are the operators of infinitesimal rotations. 2. Prove the following commutation relations: a) \h, x k \ = ie ild x\, b) [/*, p k \ = ie ild Pi, where e ikl is the antisymmetric unit tensor of third rank, the com- ponents of which change sign for any interchange of two of its indices, e Vti = — e iik> and where e 123 = l (1,2, and 3 correspond to Prove the following relations [/, (x*+y*+z 2 )] = 0. Show that in a state $ with a well-defined value of 4 Oy* = my) the average values of l x and /„ are equal to zero. Hint. Find the average value in the state ij> for the left hand side and for the right hand side of the commutation relations Urn— Uz = tiy 5. Obtain an expression for the operator of the angular mo- mentum relative to an arbitrary axis {z') in terms of the operators l X ' ly ''z' 6. Show that if in the state <]> i z y = mty the average value of the angular momentum relative to an axis z' which makes an angle with the z-axis is equal to mcosQ. This result can be visualised as follows. The angular momen- tum vector in the state ty m is evenly "spread out" o ver a cone with its axis along the .z-axis, its slant height equal to Vl(l-{- 1), and its height equal to m. The average value of its projection on the xy - plane is equal to zero, and its component along the ^'-axis is, after averaging, equal to mcos6. 7. Find the transformation of the spherical harmonics Y n , Y l0 , and Y l _ l under a rotation of the system of coordinates that is, x, v. z) 3. a) b) 4. ANGULAR MOMENTUM. SPIN 23 characterised by the Euler angles 0, <J>, and <j>. Hint. Express the spherical harmonics in the form: 8. In a Stern -Gerlach type of experiment the deflexion of a beam of atoms with total angular momentum J depends on the value of the component of the angular momentum in the direction of the magnetic field in the apparatus. If the particles in the beam have a well-defined value of the angular momentum relative to an axis which is not along the direction of the applied magnetic field, the beam is split into 27+ l components. Determine the relative intensity of these components if J= 1, and if the component of the angular momentum along an axis which makes an angle 6 with the direction of the applied magnetic field has the well-defined value M(+l, 0, —1). 9. The * -component of the electron spin is equal to +V 2 . What is the probability that its component along a direction z' which makes an angle 6 with the *-axis is equal to +V 2 or -*/ 2 ? Deter- mine the average value of the component of the spin along this direction. 10. The most general form of the spin function of a particle of spin V 2 in the "z -representation" is ^ = e ia cos§, This function describes a particle state in which the probability that the z -component of the spin is equal to +V 2 (or -V 2 ) is equal to cos*S (or sin 2 8). What would be the result of a measurement of the component of the spin in a completely arbitrary direction ? 11. The spin function has the following form in the "z- representation" : 'V\ /^ itt cosS v Is there a direction in space along which the spin component has the well-defined value of +V 2 ? If such a direction exists, find the polar angles (b, $) of its direction. Hint . Find 8 and $ from requiring the second component of the spin function to be equal to zero. 24 PROBLEMS 12. Consider a system of one kind of non -interacting particles. Let their momentum be the same and their spin be equal to V^ If these particles did not possess spin we could describe the system by a "pure case" ensemble. However, we do not know whether the spins of all the particles are parallel. Is it possible to use an experiment of the Stern-Gerlach type to determine whether this beam of particles corresponds to a "pure case" or to a "mixture" ensemble? 13. Show that the transformation operator of the components of the spin function corresponding to a rotation over the Euler angles 8, ^ and cp is of the form 7(<|>, 9, <p) = ***•««*•£ **+•• . 14. Show that if the system of coordinates is rotated over an angle $ with respect to an axis with direction cosines a, p, and T> the transformation matrix of the components of the spin function can be written as follows f = e n < M >eVrV = cos f- + 2l («„ + fs y + T i) sin *- . Hint. The Euler angles 6, <j>, and <p are connected with a, p, T , and <3> by the following relations : COS-g- = COS -g ■ cos -jOp-H), a sin ^- = sin ^- .. cos-j(cp — <V), P sin -j- = sin -j • sin -j (cp — t|>), 1 sin -g- == cos -J • sin-j (cp + <Ji). Note that «*•<-.+ « tf » + T^> is not equal to ««•"« .«<*** ■«'•*• . 15. Find the eigenf unctions of the operator *s a -\-$s 9 -\--{s M , where a 2 _j_ ^2 _j_ ^2 _ i f and show that the expansion coefficients of an arbi- trary function (J 1 ) in terms of these functions determine the pro- bability that the value of the spin component in the direction charac- terised by the direction cosines a, p , and 7 is equal to +% or -%. 16. Find the transformation matrix of the components of the spin function of particles of spin 1 corresponding to an arbitrary rotation of the system of coordinates. 17. The angular momentum of a particle is equal to j, and its z -component has its largest possible value. Determine the probability for different values of the angular momentum component along a direction which makes an angle 8 ANGULAR MOMENTUM. SPIN 25 with the z -axis. 18. A system with total angular moment J is in the state with J, = M. Determine the probability that a measurement (for instance by a Stern-Gerlach experiment) of the angular momentum component along a direction z which makes an angle 0, with the *-axis leads to the value M.' . 19. Show that if 40? is the eigenfunction of the j\ operator corresponding to an eigenvalue m, the function <i m = «-^-<^»,(o> will be the eigenfunction of the operator: "* Vw A = J x sin * c os 9 + J v sin » sin <p -J-7* cos 0, corresponding to the same eigenvalue, that is: Mm = m^ m „ Hint. Use the relations (see problem li, section 3): e -< Vy g A® = 7 g cos » -+- 4 sin &, e'^^J^e 1 ** = J x cos 9 + J y sin <p. 20. A spin y 2 particle moves in a central field of force. Find the wave function of this particle which is simultaneously an eigen- function of the three commuting operators : 21. A state of an electron is characterised by the quantum numbers, /, j, and m. Using the wave functions obtained in the preceding question, determine the possible values of the components of the orbital and spin angular momentum and the probabilities that these values are realised. Find also the average values of these components. 22 We define the direction of the spin as that direction along which the spin component has a well-defined value of +V 2 . Let this direction be characterised by the polar angles and $ , and let the state of a particle be described by a wave function <K/, 7 — /±V 2 . m) (see problem 20, section 4). It is clear that the spin direction of such a particle will, generally speaking, not be the same in all points of space. Find the relation between the angles « and $ and the space coordinates of the particle. 23. Find that wave function of a system consisting of two spin /l particles which is an eigenfunction of each of the two commuting operators, the square and the z -component of the total spin 26 PROBLEMS 24. A system consists of two particles, one with angular mo- mentum / t = 1, and the other with angular momentum l 2 = l. The total angular momentum J can in that case take on the values l-\-l, l t and / 1. Express the eigenf unctions of the operators P and J t in terms of the eigenf unctions of the square and the z -component of the angular momentum of the separate particles. 25. Denote by « t and a 2 the spin operators of two particles and by r the radius vector connecting these particles. Show that any positive integral power of either of the operators (3^2) and S l2 = -j Wz* and any product of such powers can be written as a linear combina- tion of these operators and the unit matrix. 26. Show that the operator S 12 of the preceding question can be expressed as follows in terms of the total spin operator S = j (»i+«i) and that in the case where the total spin of the two particles is equal to unity S» can be written in the form of the following 3 by 3 matrix r 20 -V3K 2> _ X /6K 2> _ 2 ] /6K 22 VSYtx Y *o 27. Show that the normalised part of a 3£> l state wave function which refers to the spin and angular dependence can be written in the following form : 4 V2tc Hint. See question 24, section 4. ANGULAR MOMENTUM. SPIN 27 28. Prove the following equations: a) {A A)=l([Af\ — [JA]) t b) {A {A A))=2(J*A + AJ*) — 4J{JA), \n'J C > WfnJM ~ 7(7+1)" GO/*- The arbitrary vector quantity A satisfies the commutation rule {Ji, A k }=le m A t . 29. Find the average value of the operator jt = g t f t + gj 2 in the state characterised by the quantum numbers J, Mj, j v and y 8 , if the total angular momentum J is equal to / = y*-|_/ 8 . Hint. Use the equations derived in the preceding question. 30. Find the magnetic moment (in nuclear magnetons) of the lfi N nucleus, for which one proton in a p lu state is missing from a closed shell. The magnetic moment of a* free proton is equal to V P = 2.79. 31. Evaluate the magnetic moment of the lr O nucleus which contains one neutron in a d % state apart from closed shells. The magnetic moment of a free neutron is equal to p n = -1.91. 32. What would be the value Of the magnetic moment of the deuteron if the deuteron is in the following states : a) «5 lt b) i/V c) 8 P lf d) »D t . 33. Assuming the deuteron ground state to be a superposition of a 8 5,, and a 3 D t state, determine the weight of the D-state, if {x p =-2.78, ^ = —1.91, fi d = 0.85. 34. Express the quadrupole moment of the deuteron in terms of the mean square distance apart, assuming the deuteron to be in (a) a ip t state, or (b) a *P t state. 35. Use the expressions for the matrix elements of vectors to show that the quadrupole moment of a nucleus is equal to Q = /(2/-,)J I (2(/ + D|«:/ /+i r-2/|«;/ J . I p) . The summation is over all Z protons, / is the nuclear spin, and n is the combination of all the other quantum numbers which charac- terise the state of the nucleus. 36. Let Oi denote the spin variable of the i-xh electron. This variable can take on the two values +1 and —1 . Show that if the operators 28 PROBLEMS which refer to the /-th electron act upon a function f(a v o 2 , .... a t , . . . , a n ) of the spin variables of n-electrons, the result is the following one: <3lxf = f(?v •••■ a l-v — a l> Q l+i' •••• °n)» <»»*/ = — ^lf(Pi a l-i> — a l> °J+l> •••• °n)» 37. Use the result of the preceding question to show that the operator of the square of the total spin moment of n electrons can be written in the form where P kl is the operator which interchanges the spin variables <s k and oj, that is = f(°l Ofc_i. 3 J> °*+l' •••' °I-1« °*» °M-1 °n)' 38. Show that in a system consisting of two spin V 2 particles the total spin S is an integral of motion, provided the Hamiltonian is symmetric in the two spins. 39. A system consists of two particles, one of spin l / 2 , and one of spin 0. Show that the orbital angular momentum is an inte- gral of motion for any law of interaction between these particles. 40. A nucleus A of spin 1 is excited in an even state. Ener- getically the emission of an a-particle is possible, A-*B + a. The nucleus B produced in this reaction is stable, has zero spin, and is also in an even state. Use the principle of conservation of angular momentum and of parity to show that this reaction is for- bidden. 41. Show that the orbital momentum L of the motion of two a-particles is always even (L = 0, 2, 4, ... ). 42. Is it possible for the |Be nucleus in an excited state with spin 1 to decay into two a-particles? 43. Use the fact that the only bound state of the neutron -proton {n, p) system is even, that the total spin of that state is unity and that the (n, n) and (n, p) interactions are identical to show that two neutrons cannot form a bound state. CENTRAL FIELD OF FORCE 29 44.* Express the operator of rotation over a finite angle cp around the direction n in terms of the angular momentum operator (of a system of N particles). 45.* Show that the equation ~M* = m(l-t-l) can be obtained by using elementary equations of probability theory, if one starts from the fact that the possible values of the components of the angular momentum along an arbitrary axis are equal to mfi (/n =—/,.. ., , .... /) and that all these components are equally probable, and all axes equivalent. 46.* Give a simple interpretation of the commutability of the operators of the components of the linear momentum and the non- commutability of the operators of the components of the angular mo- mentum, starting from the kinematic meaning of these operators, which are connected with infinitesimal translations and rotations. 47.* Show that for a system consisting of two identical particles with spin / the ratio of the number of states symmetrical in the two spins to the number of states antisymmetric in the two spins is equal to i±i. 48.* Give the normalised wave function of a system of three identical, weakly interacting bosons, which are each in a given state. 5. CENTRAL FIELD OF FORCE 1. A particle moves in a central field. Write the equation for the radial part R nl of the wave function in the form of a one- dimensional Schrodinger equation. 2. Find the radial part of the wave function of a particle in a central field using the semi-classical approximation. 3. Show that in the case of a discrete spectrum in a central field the minimum value of the energy for a given value of the orbital quantum number / increases with increasing /. 4. A system consists of two particles of mass ju t and f/ 2 . Express the operators of the total orbital angular momentum £ + /* and the total momentum p x -\-p z in terms of the centre of mass co- ordinate /?= tVi + tVa and the relative distance r = r 2 —r.. V-i + IJ-a Show that if the potential energy of the interacting particles depends only on their distance apart, U = U ( | r 2 — r x \ ), the Hamiltonian can be put in the form 2(fx 1 + (A3 ) ** 2t^— A r+U(r), 30 PROBLEMS where A r and A* are the Laplace operators referring to R and r. 5. Determine the wave functions and energy levels of a three- dimensional isotropic oscillator. 6. Solve the previous problem in Cartesian coordinates by separation of variables. Express the wave function for n r = 0, / == 1 (see previous question) as a linear combination of the wave functions obtained. 7. Assume that a nucleon in a light nucleus moves in an averaged potential of the form U(r) = — U -\-^- r 2 , to determine the number of particles of one kind (neutrons or protons) which can be accommodated in a closed shell. A shell is defined as the totality of all states with the same energy. 8. Calculate the theoretical radius of the closed shell nuclei | He and l \0 assuming the same potential as in the preceding ques- tion. The theoretical nuclear radius is defined as the distance from the centre of mass of the nucleus to the point where the "nuclear density' ' p (0 = 2 <£ (?) <|>v ( r ) (the summation is over all nucleons) decreases most steeply, that is, 9. We can write approximately U (r) = — Ae- r ' a for the inter- action between a proton and a neutron. Find the wave function of the ground state (/ = 0) and determine the relation between the well depth A and the quantity a, which characterises the range of the force, by using the empirical value E = -2.2 MeV for the deuteron energy. 10. Determine approximately the deuteron ground state energy for the potential U (r) = — - Ae-n a (A = 32 MeV, a = 2.2 f ermi) using the Ritz variational principle. Use as trial wave function a function of ar the form R = ce~^, depending on one parameter a. The value of oo c follows from the normalisation condition J R 2 r 2 dr = \. o 11. Determine the energy levels and wave functions of a par- ticle in a spherical "potential well" L/(r) = (r<a)\ U (r) = oo (r>a). Consider the case where / = 0. CENTRAL FIELD OF FORCE 31 U(r) 12. Determine the discrete energy spectrum of a particle with zero angular momentum in a spherical potential well (r > a). 13. Apply perturbation theory to determine qualitatively the change in energy levels when we change the potential ' — U (r<a) (r > a) ' * t0 the P° tential given in fig. 18. -( U(r) -I Um Fig. 18 Fig. 19 14. The potential energy of an <*-particle in a nucleus consists of two parts: the Coulomb repulsion and the short range nuclear attractive force. The total potential energy is sketched in fig. 19 The emission of an a-particle is a typical quantum effect, deter- mined by a tunnel effect. Consider the transmission of a particle of zero angular momentum through a spherical potential barrier of the following simple form £/(r) = (rO-O, U(r) = U (r 1 <r<r 2 ), */(r) = (r 2 <r). Find a relation between the lifetime and the energy. 15.* Investigate the motion of a negative muon in the field of a nucleus of charge Ze (to be considered as a sphere of radius R, with the charge uniformly spread over its volume), assuming that the forces of interaction between the meson and the nucleus are purely electrostatic of character. Find the wave functions and energy levels of the stationary states of the meson in the limiting cases of small and of very large Z. 16.* Find the selection rules for transitions between stationary states of particles in a central field of force under the influence of a 32 PROBLEMS perturbation, whose operator is proportional to the gradient opera- tor (V). 6. MOTION OF PARTICLES IN A MAGNETIC FIELD 1. Let the wave function of an electron at t = be of the form W(x, y, z, 0) = <!>(*, )'> 0)<P(*' 0). The wave function at time t in a homogeneous magnetic field Sftvfhich is along the z-axis will then also be of the form W(x, y,z, f)=<K*. y> 0Xcp(z. 0. since in the Schrodinger equation the 2-coordinate can be separated. Show that the function ty(x,y,T) regains its initial value, apart from a phase factor, if T is the period of the classical motion in the mag- netic field. 2. Show that if a magnetic field is present the velocity compo- nent operators satisfy the following commutator relations: Va°V — V* = p.2 c - * * * le h °u> . v y v z — v z v v = -^3V x , v z v x — v x v 3 = ^etf y . 3. Use the results of questions 2, section 6, and 5, section 1, to determine the energy of a charged particle moving in a constant magnetic field. 4. Determine the energy spectrum of a charged particle moving in a uniform magnetic field and a uniform electric field which are at right angles to each other. 5. Determine the wave functions of a charged particle in a uni- form magnetic and a uniform electric field which are at right angles to one another. 6. A charged particle moves in a homogeneous magnetic field and in a central field of the form U(r) = J_±_ . Determine its energy spectrum. 7. Determine the time dependence of the coordinate operators jc and y of a charged particle in a uniform magneti c field (vector potential A x ^ — ^y,A v = ^-x, A z = 0). Find (x—xf and (y — yf as functions of time. MOTION OF PARTICLES IN A MAGNETIC FIELD 33 8. Determine the energy levels and wave functions of a charged particle in a uniform magnetic field. Use cylindrical coordinates. Write the vector potential in the form A 9 = -&p, A f = A g = 0. 9. Find the current density components for a particle in a uni- form magnetic field in the state characterised by the quantum num- bers n, m, k z (see preceding question). 10. Find in cylindrical coordinates the energy levels of a charged particle in a uniform magnetic field using the semi -classical approximation. 11. Determine the classically accessible region of the radial motion of a particle in a magnetic field (see preceding question). 12. Estimate the minimum "spreading out" in the radial direc- tion of the orbit of a charged particle in a magnetic field. 13. Use classical mechanics to express the coordinate of the centre of the circle along which a charged particle in a uniform magnetic field moves in terms of the coordinates x, y and the gen- eralised momenta p x and /y Consider the coordinates and momenta in that expression to be operators and find the commutation relation for the coordinates of the "centre of the orbit" introduced in this way and the corresponding Heisenberg relation. Show that the sum of the squares of the coordinates of the "centre of the orbit" takes on the discrete values 2fic, . , v where 1,2,... 14. Show that in a variable uniform magnetic field the wave function of a particle with spin can be written as the product of a space function and a spin function. 15. A spin V2 particle is in a uniform magnetic field along the *r-axis, the absolute magnitude of which varies arbitrarily §6 — 38 if). The spin function at t = is of the form g-* a cos8 e ia sin 8 Determine the average value of the x- and y -components of the spin and also the direction in space along which the spin at time t has a well-defined value. 16. In the half-space x > there is a uniform magnetic field <&'a> = 3@ v = 0. 36 z = §6, while there is no field in the half-space x < . A beam of polarised neutrons of momentum p falls on the plane x = 0, coming from the region x < . Find the reflexion coefficient of the neutrons at the dividing boundary. 34 PROBLEMS 17. A spin V 2 particle is in a uniform magnetic field, the abso- lute magnitude of which is constant and which varies in time accor- ding to the equations gf6 x =z gff> sin cos wt, <§e y = S/C sin ft sin u>/, ^ e = J£cos&. At f = the spin component in the direction of the magnetic field was equal to + 1 / 2 - Determine the probability that at time t the par- ticle has made a transition to the state where the spin component along the magnetic field direction is equal to -V 2 . 18. A spin y 2 particle with a magnetic moment p. is in a non- uniform magnetic field of the form &e z = <ffi Q -\-kz, SV y = — ky, 3V X = (div$? = 0). a) Find an expression for the time dependence of the x, y, z coordinate operators. b) Determine the average values of the coordinates and the time dependence of their dispersion, assuming the particle wave function at t = to be of the form ipox ¥ = *(*. y. z)e * (») 19. A neutral particle is in a spatially uniform magnetic field; the direction of this field, but not its absolute magnitude, varies with time. Write down an equation for the spin function in the n |- representation" where the £-axis is along the direction of the magnetic field. Show that if the variation in the direction of the magnetic field is sufficiently slow, the probability for different values of the component of the spin along the magnetic field direc- tion will not change in time. 20.* Find the conditions under which the Hamiltonian of a charged particle in a magnetic field is of the form o-W-W *=£*-£#+•&*• ATOMS 35 7. ATOMS 1. Use the inequality to find the minimum energy of a one-electron atom and the corres- ponding wave function^ Show that for the ground state of the atom the inequality 2r>jK| is satisfied. 2. An electron moving in the Coulomb field of a nucleus of charge Z is in its ground state. Show that the average electrostatic potential in the neighbourhood of the nucleus and the electron is givei by the equation 17 r HZ— i) , /z . i\ ~ i m\ 3. Show that in the ground state of the hydrogen atom a) the most probable value of r is equal to a ! ' 2 b) the average value — = — , c) !=-?-. 4. The wave function <|>(r) describes the relative motion of two particles, a proton and an electron. Let the coordinates of the centr of mass of the hydrogen atom be accurately known and be equal to X=0, Y = Q, Z = 0. Show that in this case the probability density for the proton is of the form ixe* '">=(^ri*(^)r. where m and M are, respectively, the electron and the proton mas; 5. Find the momentum distribution of the hydrogen atom elec- tron in the Is, 2s, and 2p state. 6. Evaluate r 2 — r 2 f that is the dispersion of distance of the electron from the nucleus for the hydrogen atom in the state with quantum numbers n, I. 7. Express the hydrogen atom eigenf unctions in parabolic co- ordinates corresponding to n^l, b 2 = 0, m = in term s of the wave functions in spherical coordinates. Show also that 1V = 0, j?,=0, m = n~l(Z, 1\, <?) = tyn,l = n-l,m = n-l(r, 0, cp). 36 PROBLEMS 8. Show directly that the degree of degeneracy of the n-th hydrogen atom energy eigenvalue is equal to n 2 by considering the solutions of the SchrOdinger equation in parabolic coordinates. 9. Find the correction to the hydrogen atom energy levels taking into account the relativistic change of mass with velocity (consider the terms of order -^ ) . 10. From the relativistic equation for the electron (Dirac equa- tion) it follows that apart from the correction following from the change of mass with velocity (~— ) there is also a term "2 — 2fA 2 C a r dr lS ' in the Hamiltonian, where /is the orbital angular momentum opera- tor, s the spin operator, and U(r) the potential energy of the elec- tron which we assume to be spherically symmetric. The physical meaning of this term consists of the fact that the movement of a mag- netic moment n (connected with the electron spin) leads to an elec- trical dipole moment </ = — [t>{t], which interacts with the field of the nucleus. Find the correction to the hydrogen atom energy levels by taking the term H 2 (the so-called spin-orbit interaction) into account. 11. Show that the quadrupole moment of the hydrogen atom is equal to ; y+i 2 -p i — n* {*=T< 5fl H-l-3f (/+!)(£)' 12. Determine the total probability for excitation and ionisation of the tritium atom 3 H when its /S-decay takes place. Calculate also the probability for exciting the n-th level. 13. Use the variational method to find the ground state energy of a two-electron system in the field of a nucleus of charge Z. Use as trial functions a product of hydrogen-like wave functions with effective charge Z'. Neglect relativistic corrections. 14. To a fair approximation we can put for the wave function of a helium atom the following expression: 3 Z'(r 1+ r,) ^ = ±-^e " lz'=~\ (see problem 13, section 7). Show that the electrostatic potential around the atom is of the form: *<'> = 2 '(t + t)«' 2Z'r ATOMS 37 15. Evaluate the diamagnetic susceptibility of helium using the approximate wave function of its ground state (see problem 13, sec- tion 7). 16. Use the variational method to determine the energy of the ground state of a lithium atom taking exchange into account. Take for the eigenfunctions of the electrons the following hydrogen -like functions: for the Is electrons a function of the form <|; 100 == 2Zi'e~ z > r , and for the 2s electron a function of the form -?•£. *? m = cZh 2 (1— tZ z r), where Z, and Z 2 are variational parameters, where c follows from the normalisation of # 2 oo> anc * V from the fact that ^ 100 and ^ 2 oo are orthogonal on to one another. If we apply to this problem ordinary per- turbation theory, we put Z 1 = Z 2 — 3. By introducing Z t and Z 2 as variational parameters, we include in our considerations the screen- ing effect of the electrons. 17. Determine the shift of the atomic energy levels due to the motion of the nucleus. Evaluate this shift for helium in the triplet and in the singlet ( Is ) ( np ) state, using the eigenfunctions in the form of hydrogen -like functions with an effective charge for the dif- ferent electrons. 18. Let the potential energy U (x, y, z) be a homogeneous func- tion of the coordinates of degree of homogeneity v , U(lx, Xy, X2) = M/(*, y, z). Show that the average value of the kinetic energy in a state of the discrete spectrum is connected to the average value of the potential energy by the relation 2T = -iU (virial theorem). 19. Estimate the order of magnitude of the following quantities using the Thomas-Fermi model: a) average distance of an electron from the nucleus; b) average Coulomb interaction energy of two electrons in the atom; c) average kinetic energy of an electron; d) energy necessary to ionise the atom completely; e) average velocity of an electron in the atom; f ) average angular momentum of an electron; g) average radial quantum number of an electron. 20. Use the Thomas-Fermi model to obtain an approximate expression for the energy of an atom in terms of the electronic 38 PROBLEMS density p (/")*. 21. Show that one obtains the Thomas-Fermi equation by mini- mising the total energy with respect to the density p(r). Hint. Use the results of the preceding question. In varying £(p) one must take into account the normalisation condition pdx = N (in the case of neutral atoms N = Z). S 22. Use variational methods to find the best approximate expression for the electronic density in the Thomas-Fermi model, Ap" x / ~r using trial functions of the form p = —— , x=y y , where A follows from the normalisation condition J pdx = N (N = Z for neutral atoms) and where X is a variational parameter. Find the energy of the atom (or ion). Note. In determining the form of the trial functions, we took into account the fact that the exact solution must for small values of r be of the form const 23. Prove the virial theorem for the Thomas-Fermi model. 24. Use the virial theorem to show that in the Thomas-Fermi model for the case of a neutral atom the electrostatic interaction energy of the atoms is one-seventh of the electrostatic interaction energy between the electrons and the nucleus. 25. Evaluate in the Thomas-Fermi approximation the energy needed to ionise an atom (ion) completely. 26. Determine the shift of the energy levels due to the finite size of the nucleus. Assume the potential within the nucleus (r < a) to be constant (this corresponds physically to a distribution of the nuclear charge over the surface of a sphere of radius a). 27. Evaluate the value of ^ 2 (0) for a valence s -electron in an atom of large Z, using the semi -classical approximation. 28. Determine the contribution to the magnetic field strength at the centre of a hydrogen atom due to the orbital motion of the elec- tron. Evaluate this quantity for the 2p state. 29. What is the change in the expression for the magnetic mo- ment of the hydrogen atom if the motion of the nucleus is taken into account ? * In problems 20 to 25 the units are chosen in such a way that <? = & = ATOMS 39 30 . Determine the separation of the hyperf ine structure term s for an s -electron in the hydrogen atom. 31. Determine the energy of the hyperf ine structure in a one- electron atom with non -vanishing angular momentum. 32. A diamagnetic atom is placed in an external magnetic field. Determine the magnitude of the induced magnetic field strength at the centre of the atom . 33. Solve the preceding problem in the case of helium - 34. Give the possible values of the total angular momentum in a 1 S, 3 5, 3 P, 2 D, or 4 D state. 35. Which states (terms) can be realised for the following two electron systems: a) nsn's, b) nsn'p, c) nsn'd, d) npn'p. 36. Give the possible terms of the following configurations: a) (tip)*, b) (ndf, c) ns(n'p)*. 37. Determine the lowest term of the following elements: O, CI, Fe, Co, As, La. For the electronic configurations of these atoms see, for instance, Semat: "Introduction to Atomic and Nuclear Physics", 1954, pp. 248-250. Hint. It is necessary to use the following empirically established rules. a) The lowest energy corresponds to the terms of the largest value of 5 for a given configuration of the electrons and to the largest value of L possible for this value of S (Hund rules). b) The ground state of the atom corresponds to J=\L — S\ if the unfilled shell is less than half full, and to J = L-\-S if it is more than half full. 38. Determine the parity of the ground states of the following elements: K, Zn, B, C, N, O, CI. 39. A. system of Af electrons is characterised by N triples of quantum numbers n, I, m z . Determine the number of states corres- ponding to a given value Ms of the z -component of the total spin. 40. Find the number of states corresponding to the configuration nf. 41. Show that if at<2/4- 1, the term with the largest value of L for the nf configuration will be a singlet term with L = xl — — Lx(x — 2) if a: is even, and a doublet with L =*/ — ~(x — l) 2 , if x is odd. 42. Construct the eigenf unctions corresponding to a p s con- figuration and characterised by the quantum numbers S, L, M s , 40 PROBLEMS M L , using the wave functions of the one -electron problem. Hint. Consider the action of the operators (L x — il y ) and (S x — iS y ) on the zero-th order antisymmetric function. 43. Obtain the eigenf unctions for each of the two 2 D terms of the d z configuration. 44. Two electrons move in a central field. Consider the elec- trostatic interaction between the electrons as a perturbation. Find the first order energy shifts for the terms of the npn'p configuration. Hint. The sum of the roots of the secular equation is equal to the trace of the secular determinant. 45. Show that the spin -orbit perturbation given by the equation Vsl = ASL, is such that the average perturbation of all states corres- ponding to a given term (which is characterised by given L and S) is equal to zero. 46. Find the shift in the energy levels of an atom in a weak mag- netic field when — &€ <C| Afij-j- 1, where LE jj< is the distance between the levels of a multiplet. 47. Find the range over which the Lande g--factor can change for given values of L and S. 48. Show that the terms 4 D, /2 , *F V p Oa/ 2 show a shift which is linear in the field. 49. Determine the Lande factor for a one -electron atom (hydro- gen, alkali metals) directly using the Pauli eigenf unctions (see problen 20, section 4). 50. Express the magnetic moment of an atom in terms of the Lande factor. 51. Determine the shift in the term levels of a one -electron atom the case of an intermediate field ^- &e ~ | A% |. 52. Find the wave functions of an electron under the conditions of the preceding question. 53. Determine the shift in the hydrogen energy levels in a strong magnetic field [~ £?e > | E nlf -E n i y | ) . To apply perturbation theory it is necessary to require that the energy of the atom in the magnetic field is small compared to the difference in energy between different multiplet s, that is, I eh&e ^ | P P I 2u, c "^ \ c nlj En'lj !• ATOMS 41 54. Determine the Zeeman splitting of the components of the hyperfine structure term 2 5 V; , (j = y 2 , I = 0) for the case of an inter- mediate magnetic field (~ $c ~ | AE fr | ). (The splitting produced by the field is of the same order of magnitude as the hyperfine structure splitting). 55. Show that the sum total of the changes produced by an arbi- trary magnetic field in the energies of all the states of a given value of Mj is equal to ^^^{■+ J(J+ ' ) -^?) +5(5+ " }- The summation is over J, satisfying the conditions L-f-S<y< <-\L — S\, J>Mj. 56. Show that for a hydrogen atom in a uniform electric field a) the energy of a state with / = n — 1 , m = n — 1 does not change to the first order in the field; b) the position of the centre of gravity of the displaced term is not changed; c) states which differ only in the sign of the z -component of the angular momentum have the same energy. 57. Evaluate the shift of the hydrogen atom energy levels in a weak electric field (Stark effect small compared to the fine structure splitting). 58. Find the magnetic moment of a hydrogen atom in a weak electric field. 59. Evaluate the shift of the n = 2 hydrogen atom term in an electric field of intermediate intensity (Stark effect and fine struc- ture splitting of the same order of magnitude). 60. Consider an atom acted upon by a perturbing potential a. Using perturbation theory we obtain for the wave function ip in first approximation an expression of the following form: and for the energy To apply now the variational method we simplify as far as possible the form of ip. Since CO CO 2 "no'h = — "ool'o + 2 "notn = % (" ~ "oo)» 42 PROBLEMS we use the following approximate form for ip : ♦ »*,(! +ii>). where E' in many cases can be considered to be equal to the average value of E — E n . Since the perturbed wave function is now approxi- mately determined, we can apply the variational method to determine the energy. Use the variational method to determine the energy of an atom acted upon by a perturbing potential a. Look for the minimum energy using a trial function of the form ^ = t^ (l -\-ku), where A. is a varia- tional parameter. 61. Use the result of the preceding question to find a formula for the polarisability of an atom. Give a numerical estimate of the polarisation of a hydrogen and of a helium atom, in their ground states. 62. A hydrogen atom is in a parallel magnetic and electric field. Find the energy level shift in the following cases: a) weak fields (Stark effect and Zeeman effect small compared to the fine structure splitting); b) intermediate fields, for the terms of principal quantum num- ber n = 2. 63. A hydrogen atom is in an n = 2 state in mutually perpendi- cular electric and magnetic fields. Determine the energy level shift assuming that the fields are strong (energy of the electron in the ex- ternal electric and magnetic fields large compared to the fine struc- ture splitting). 64.* Evaluate the average value of the n-tb. power of the radius r and also its dispersion for the ground state of the hydrogen atom. 65.* How does the probability of the capture of a negative meson, which is in the K -orbit of a mesonic atom, depend on the charge Z of the nucleus? 66.* Find the effective (average) potential <p acting on a charged meson passing through a non-excited hydrogen atom (whose polarisa- tion may be neglected). Obtain the limiting expression for <p for large and small distances of the meson from the nucleus. 67.* Show that the average value of the dipole moment of a sys- tem of charged particles which is in a state of well-defined parity is equal to zero. 68.* Show that if a system of N charged particles moves in a finite region of space, the following relation (the so-called "sum- rule") holds: ATOMS 43 n where rf OTn is the matrix element of some component of the dipole moment of the system and where the summation extends over all states of the system, while |i and « are the mass and charge of each particle. 69.* A plane rotator with moment of inertia / and electrical di- pole moment d is placed in a uniform electric field 8, in the plane of rotation. Consider g to be a perturbation and evaluate the first non- vanishing correction to the energy levels of the rotator. 70.* A three-dimensional rotator with moment of inertia / and electrical dipole moment d along the axis of the rotator is placed in a uniform electric field 8 which is considered to be a perturbation. Evaluate the first non-vanishing correction to the ground state ener- gy of the rotator. 71.* Evaluate in first order perturbation theory the ground state energy of a two -electron atom or ion with nuclear charge Z, taking the interaction between the electrons as a perturbation. Evaluate also the first ionisation potential of the atom (ion) con- sidered. 72.* A linear harmonic oscillator is acted upon by a uniform electric field which is considered to be a perturbation and which depends as follows on the time: where A is a constant. (Since the action of a uniform field is equi- valent to a shift of the point of suspension, this problem can be solved not only by perturbation theory, but also exactly, using the methods of problem 44, section 3. ) Assuming that when the field is switched on (that is, at * = ~oo) the oscillator is in its ground state, evaluate to a first approximation the probability that it is excited at the end of the action of the field (that is, at.t = + oo). 73.* Solve the preceding problem for a field which varies as follows Sr/)~_ !__ and which corresponds to a given total classical imparted impulse P. 74.* A uniform periodic electric field acts upon a hydrogen atom which at * = is in its ground state. Determine the minimum frequency of the field necessary to ion- ise the atom and use perturbation theory to evaluate the probability for ionisation per unit time. For the sake of simplicity assume the 44 PROBLEMS electron in the final state to be free. 75.* Find the probability for the ejection of a /C-electron from an atom accompanied by a dipole transition of the nucleus assuming direct electrostatic interaction between the electron and the protons of the nucleus (internal conversion, neglecting retardation). Use the wave functions of a /C-electron in a hydrogen-like atom for the initial wave function, and assume the velocity of the electron in its final state to be much larger than atomic. 76.* The nucleus of an atom in a stationary state ^ experiences a sudden impact during a time * which imparts to it a velocity v. Assume that x <C 7* and* <C — , where T and a axe of the order of magnitude of the electronic periods and the dimensions of the elec- tron shells, and express in general form the probability that the atom has made a transition to a state ty n as the result of such a "kick". 77.* Use the result of the preceding problem to evaluate the total probability of excitation or ionisation of a hydrogen atom (initially in its ground state) as the result of a sudden M kick M during which the proton receives a momentum p. Discuss the conditions for the applicability of the result obtained. 78.* The usual quantum rule of the semi -classical approximation is obtained for the case where the region in which the particle can move, is bounded by two turning points, in the neighbourhood of which the semi-classical approximation cannot be applied. Another case is met with when one considers the motion of electrons in the Thomas- Fermi distribution which are in an 5 -state. (The same result applies to the motion of electrons in any central field of force which at small' distances from the centre goes over into a Coulomb field; the Thomas- Fermi model is only taken to fix our ideas. ) The region in which these electrons can move is bounded for small values of r by the point r = 0, which is not a turning point. On the other hand it is wrong to demand that the semi -classical function is finite at r = o, since in the neighbourhood of this point the semi-classical approximation cannot be applied. Obtain the quantum rule for s-electrons of the Thomas-Fermi distribution. 79.* Estimate the order of magnitude of the polar isability of a Thomas-Fermi atom, that is, the ratio of the dipole moment d of the Thorn as -Fermi electron distribution, produced by the action of an applied electric field, to the value of the electric field strength $ . 80.* Investigate the influence of the finiteness of the nuclear mass M on the energy levels of an atom with n electrons, assuming that the enerev levels are known for M = <x>. MOLECULES 45 81.* Evaluate the "exchange" correction (that is, the correction determined by the Pauli principle) to the terms of a two-electron atom (or ion), assuming one of the two electrons to be in the ls-state, neglecting the electrostatic interaction between the electrons and taking into account the fact that the nuclear mass is finite. 8. MOLECULES 1. Obtain the SchrOdinger equation for a diatomic molecule assuming that the centre of mass of the molecules is approximately the same as the centre of mass of the nuclei. To describe the motion of the electrons use a moving system of coordinates fixed at the nuclei. Spin effects can be neglected. 2. Solve the preceding problem by considering the spin states of the electrons and describing these states in the moving system of coordinates 5, -q, r. 3. For small vibrations of the nuclei the wave functions of a molecule can be approximated by a product of three functions #el (ii, rii, Cj, a it p), /(p), ft (8, <p). The first function describes the motion of the electrons while the nuclei are fixed, while the second and third describe the vibrational and rotational states of the molecule. Find the equations which determine the vibrational and rotational parts of the wave functions of diatomic molecules. 4. Determine the possible terms of the diatomic molecules N 2 , Br 2 , LiH, HBr, and CN, which can be obtained by combining the two atoms in their ground state. 5. Find an equation which determines the electronic terms of a hydrogen atom interacting with a helium atom, assuming both atoms to be in their ground state. 6. Find the vibrational and rotational energy spectrum of a di- atomic molecule, taking into account the fact that the nuclei move in a potential of the form V(r) = -2D(i- 2 i), where P = £. 7. Express the effective potential of the preceding question V(r)-f- 2~p2^'(*'H"- 1 ) near its minimum in the form of an oscillator potential and find the energy levels for small vibrations. 8. Determine the moment of inertia and the internuclear dis- tance of the 1 H 35 C1 molecule, if the difference in the frequency of 46 PRODLKMS two neighbouring lines of the rotational -vibrational (infrared) band of ^Cl is equal to A-.—: 20.9 cm' 1 . Evaluate the corresponding A-. for the DC1 spectrum. 9. Evaluate the ratio of the difference in energy of the first two rotational levels to the energy difference of the first two vibrational levels for the HF molecule. The moment of inertia of HF is equal to / =1.35 x 10~ 40 g x cm 2 and the vibrational frequency is equal to Ar vib = 3987 cm -1 . 10. Determine the dissociation energy of the D 2 molecule, if the dissociation and zero point energy of the H 2 molecule are res- pectively equal to 4.4G eV and 0.2(5 eV. 11. One often uses for the approximate form of the potential energy of a diatomic molecule a Morse potential V — D(l — £~ 2|,; ) 2 ; f=r r -^-^ . Determine the vibrational energy spectrum for K = 0. 12. Show that the operator of the square of the total angular momentum of a diatomic molecule can be written in the form *— {is(-»«»)+^(*-'*«»<'J , l+ Ai? - 13. The £, 7], C axes are axes of an orthogonal system of co- ordinates which is rigidly connected with a rotational rigid body. Find the operators J iy 7 V j r of the \, tj. C components of the angular momentum of the rigid body. 14. Show that the operators .A, i Tj , X obey the following com- mutation relations J- 7 T| 7 f| 7- = — iJ^, J T Jr— ll, =— Ui, which shows that the commutation relations for the angular momen- tum components in a moving system of coordinates differ from the commutation relations in a non-moving system only in the sign on the right hand side of the equations written down. 15. In classical mechanics we have the Euler equations: Af t +{C-H)qr = 0, B d -± + {A — C)rp = 0, MOLECULES 47 or l f + (F-^)V— 0,etc. Show that in quantum mechanics the last relations are of the form ^ + 7(i'~"e r ) ( V' + A^) =0, etc. 16. Molecules with two or more axes of symmetry of third or higher order (for instance, CH 4 ) represent a spherical top. For such molecules the inertial ellipsoid goes over into a sphere .4 == R = C. Determine the energy levels of a spherical top. 17. Molecules with an axis of symmetry of third or higher order (for instance, S0 2 . NH 3 , CH 3 Cl)and molecules with lower symmetry or even without any symmetry, but of which two of their principal moments of inertia are equal can be considered to be symmetrical tops, A = B4C. Determine the energy levels of a symmetrical top. 18. Give the SchrSdinger equation of a symmetrical top. 19. Find the eigenf unctions of the operator >=■,_{ J-ifsinO-^-U-i-f-^-l-^— 9 ( i°- s l di \ I sin dO \ dQJ ' sin* \d^ ~*~ d^J ~ sin- 0? di I " 20. Evaluate the matrix elements of the Hamiltonian of an asymmetric top. 21. Determine the energy levels of an asymmetric top for y= 1. 22. Find the wave functions of an asymmetric top for 7=1. 23. Use the properties of the Pauli matrices to show that even 'if spin-spin interactions are taken into account the 2 S terms of a diatomic molecule are not shifted. 24. Determine the multiplet shift of a 3 2 term related to a Hund's b -type bond. 25. When the SchrOdinger equation is approximately solved (see problem 3, section 8) one does not take into account the operator since its diagonal elements are equal to zero. A consideration of the off -diagonal elements related to the same electronic (n, A) and vi- brational (v) state leads to an effect, the so-called rotational distor- tion of the spin. Considering the operator to be a perturbation, * Landau and Lifshits, "Quantum Mechanics", Pergamon Tress, 1958, p. 282. In this case, the sjiin-orbit interaction energy is small compared to the energy dif- ference between rotational levels. 48 PROBLEMS determine the change in the doublet levels due to this perturbation. 26. Derive a connexion between the value of the total nuclear spin angular momentum of the D 2 molecule in a 2 -state and the possible values of the quantum number K. 27. Determine the Zeeman shift of a diatomic molecule term, if the term is a case a term. The magnetic field is assumed to be small, that is, the interaction energy of the spin with the external magnetic field is small compared to the difference in energy of con- secutive rotational levels. 28. Determine the Zeeman shift of a diatomic molecule term if the term is a case b term and the magnetic field is such that the interaction energy of the spin with external magnetic field is small compared to the spin -molecular dipole interaction energy. 29. Solve the preceding problem for the case where the spin- molecular dipole interaction energy is small compared with the energy shift produced by the external magnetic field. 30. Determine the Zeeman shift of a diatomic molecule doublet term (case b) in a magnetic field of such a magnitude that the inter- action energy of the magnetic moment with the field is of the same order of magnitude as the spin -molecular dipole interaction energy. 31. Determine the shift of a diatomic molecule term (case a) in an electric field, if the molecule possesses a constant dipole moment p- 32. Solve the preceding problem for a case b term. 33. Determine the energy of a rigid dipole p in a uniform elec- tric field $, which may be considered to be a small perturbation. Hint. Use the relation fiD _i/~( H- 1 ) 2 — m2 n , ,/* P — m* p cosny lm —y (2 /H-3)(2/+l) * +1 > nr1 r (2/+l)(2/— l) , - 1 '" r 34. Use perturbation theory to determine the interaction between two non -excited hydrogen atoms at a large distance R apart. 35. Consider a system of atoms with spherical charge distri- butions. We showed in the preceding problem that two such atoms at large distances apart interact by a so-called dispersion force. The dispersion law is a quantum phenomenon and in contradistinction to the classical polarisation force it is additive. Show that the inter- action energy of two such atoms is independent of the presence of other similar atoms, that is, show that the interaction energy of a system of atoms is additive, that is, equal to the sum of the inter- action energies of different pairs of atoms. SCATTERING 49 36.* Estimate for the case of diatomic molecules the relative order of magnitude of the following quantities: a) the intervals be- tween the electronic, vibrational and rotational levels; b) the inter- nuclear distance and the zero point amplitude of the nuclear vibra- tions; c) the characteristic periods and velocities of the electronic and nuclear motion. 9. SCATTERING 1. Find the cross-section for the scattering of low velocity par- ticles by a potential well (de Broglie wave length large compared to the well dimensions). 2. Determine the cross-section for the scattering of slow par- ticles by the repulsive potential U(r) = U (r<a), U(r) = (r>a). 3. Determine the first three coefficients of the expansion in Legendre polynomials of the elastic scattering cross-section ^ in terms of the phase shifts. 4. Find the phase shifts in the field U = 4- • Determine the scattering cross-section at smali angles. 5. Evaluate the differential scattering cross-section in a re- pulsive field U= — both in the Born approximation and according to classical mechanics. Determine the limit of applicability of the formulae obtained. 6. Find the discrete levels for a particle in an attractive po- r tential £/(/-) = — £/ <T«" for / = 0. Determine the phase shift 6 for this potential and analyse the connexion between <5 and the discrete spectrum. 7. Show that in a Coulomb field there exists a one-to-one rela- tion between the poles of the scattering amplitude and the discrete energy levels. Hint. Use the equation e™i = . 50 PROBLEMS 8. Determine in Born approximation the differential and the total scattering cross-sections for the following potentials: p -<tr &)Uir) = g* e — , b) u^^Uue-*'*, C) U(r) = U (t c-* r . 9. Use the Born approximation to find the differential and the total elastic cross-section for scattering of fast electrons by: a) a hydrogen atom; b) a helium atom. 10. Consider a collision of two identical particles with inter- action energy U(r). Find the effective scattering cross-section for slow identical particles in the case of a short-range force. 11. Evaluate the cross -section for elastic scattering of an electron by an electron or of an a -particle by an d*-particle. 12. The scattering of neutrons by protons depends on the total spin of the neutron and proton. At small energies the cross-section for the triplet state ( v s= 1) is equal to cr tr =4u|/ u | 2 ^2 • 10" 24 cm 2 and for the singlet state (5 = 0) a si = 4* \f x \ 2 ^ 78 • 10~ 24 cr 2 . Introduce the operator It is easily seen that its eigenvalues are /, and f x for the triplet and singlet state respectively. To determine the scattering cross -section for arbitrarily polarised neutrons we must average the operator / : ', a = 4ic/ 2 . Let the spin state of the incoming neutrons be described by the func- /«-' a cos.3\ . . , , tion ' (the direction of the neutron spin m polar angles is \« ,a smp / ft = 23, <I>r •■iz-r--., see problem li, section 4) and that of the proton by the function ( ) (proton spin parallel to the 2-axis). Determine for this case the cross-section for the scattering of neutrons by pro- tons. 13. Find the probability that the spin of a slow neutron changes its direction when it is scattered by a proton if the neutron spin be- fore the collision was in the direction of the z-axis and the proton spin in the opposite direction. *14. Slow neutrons are scattered by molecular hydrogen. If their energies are appreciably less than thermal (ft > 10~ 8 cm, that is, the wave length of the incoming neutrons is large compared to the distance apart of the protons in the molecule) their scattering SCATTERING 51 amplitude is equal to the sum of the amplitudes due to each of the protons. Thus we have J 2 1 1 \*n (»/>, "+- »p.) J • The spins of the two protons in a hydrogen molecule can be parallel (orthohydrogen) or antiparallel (parahydrogen). 'Determine the cross-section for the scattering of neutrons by para- or ortho -hydrogen. 15. Determine the total cross-section for elastic scattering of fast particles by an impenetrable sphere of radius a (de Broglie wave length %<^Za). 16. Show the following properties of the scattering amplitude for E -+ (scattering length): a) in a repulsive potential the scattering length is negative; b) in an attractive potential, the scattering length is positive, if there are no discrete levels; c) the scattering length tends to infinity if on deepening the potential well a new level appears. Hint. If we go over from the Schr&dinger equation to the equi- valent integral equation (E — 0) T 2-n.li* J \r — r'\ we find for the asymptotic behaviour of ip where the scattering length is expressed in terms of the potential energy and '\>e^q. 17. Consider the elastic scattering of spin V 2 particles by scalar (spin zero) particles. Find the differential scattering cross-section for oriented spins. Consider the scattering of S and P waves. 18. Show that in the general case of inelastic scattering, the following formula, which connects the total scattering cross-section v — o e ] + Cinei and the elastic scattering cross-section for ft = 0, holds, k a = £lm/(0). Hint. Use the expansion of these quantities in terms corres- ponding to different values of the orbital angular momentum 52 PROBLEMS ^ a >=2k2 (2/ + 1)( ^- 1)P ' (cos{)) ' 1 = .oo oo 1=0 19. Consider as a limiting case a so-called "black" nucleus. Let the nuclear radius R be large compared to the neutron de Broglie wave length. Assume that all neutrons hitting the nucleus are ab- sorbed. Determine the total scattering and the total absorption cross- sections. 20. Consider the nuclear reaction A-\-a-+C-+B-\-b. What will be the angular distribution of the products of this nuclear reaction in the centre of mass system, or, what is the same, in the system in which the compound nucleus is at rest, for the following three cases: a) the spin of the compound nucleus is equal to zero; b) the orbital angular momentum of the relative motion of the products of the reaction is equal to zero; c) the orbital angular momentum of the relative motion of the reacting particles is equal to zero (the spin of the compound nucleus different from zero) ? 21. Find the eigenf unctions of the operators of the square and of the £ -component of the isotopic spin, P and I zi for the nucleon- meson system (see Appendix II). 22. In the centre of mass system the scattering of mesons by nucleons is the problem of scattering of particles by a stationary scattering centre. Far from the scattering centre the incoming wave with well-defined values of the 2-components of spin S z and of iso- topic spin x z can be written in the form ***"(o) 8(,c — *t>*(n — t,). The quantity ^ can take on the values: 1^=1 for a u+ meson g^ i)— cp + ., ic t = for a u° meson 8 (it) = <p , iti== — 1 for a K~ meson S(ir -|— 1) = <p_ , SCATTERING 53 while x z takes on the values: T « = -j for a proton 8 [n — -^ J = ty p , T * = — 2" for a neutron 8 (n -\- i-) = ^ n . Expand the incoming wave in terms of the eigenf unctions of the operators J 2 , J g , P, l g . Hint. In the meson-nucleon system the parity ( — 1)*, the total angular momentum J, and the isotopic spin / are integrals of motion. Since the total spin in such a system is equal to V it this will also be conserved and also the orbital angular momentum. If we ex- pand the incoming wave in terms of the eigenf unctions of operators, the absolute magnitude of which is conserved, we can sum over J and / instead of over / and /. 23. Express the scattering amplitude of pions, scattered by nucleons in terms of phase shifts for the following three cases: TZ + -\-p —*■ -K + -\-p, K~ -f- P — > It" + P, Tt~ -j-p->1t° + /l. 24. Show that the scattering amplitudes for all possible meson- nucleon reactions can be expressed in terms of those considered in the preceding problem assuming isotopic invariance. Express these quantities in terms of the scattering amplitudes for the states with isotopic spin 3 / 2 an d V%. 25. Express the total cross-sections for the following reactions in terms of phase shifts: U+-(-/7->ir+-|-/7, it- -f-p -».*--+- p, v~ H-p->it°-)-n. 26. In the region of low energies when the meson wave length is large compared with the range of the meson-nucleon forces, the main contribution to the scattering comes from 5- and P-waves only. Find the differential scattering cross-section in terms of phase shifts for the following reactions: 7c+-\-p^v:+-\~p, 7t--f-p->1t0_|_„ # 54 PROBLEMS 27. A pion beam is scattered by a target of unpolarised protons, that is, the number of protons with S s = % in the target is equal to the number of protons with S z = -%. It turns out that the protons which were originally unpolarised are polarised in the scattering pro- cess. Determine the magnitude of the proton polarisation consider- ing only 5- and P- waves. 28.* Evaluate the effective cross -section for elastic scattering by the potential U{r) = A?- r in the Born approximation. Discuss the applicability of this approxi- mation. 29.* Evaluate in the Born approximation the effective cross- section for scattering by a "delta -function" potential. 30.* Show that the scattering amplitude of a particle in an arbi- trary external field is connected with its wave function ^ through the equation 31.* A spherical rotator (that is, a particle moving on the sur- face of a sphere of given radius a) is hit by a particle which interacts with the rotator by Coulomb forces. Use perturbation theory to evaluate the differential cross-section for the inelastic scattering over an angle 6 with the simultaneous excitation of the l-th level of the rotator. 32.* Use the principle of detailed balancing to connect the effec- tive cross-section for the radiative capture of a neutron by a proton with the photodissociation of the deuteron. 33.* A system consisting of a negative and a positive pion is in a state with a well-defined orbital angular momentum I. Find the selection rule for the process it--f-ic+ ->Tr°-(-Tt with respect to whether / is even or odd. 34.* Show that the process is forbidden for a scalar pion (that is, a pion with spin and even parity) when it is captured in the S-state of "me so -deuterium". 35.* Show that it follows from the experimental fact that the process SCATTERING 55 has a very small probability that the negative and the neutral pion possess the same internal parity, assuming that the negative pion capture leads to the 5- state of "meso -deuterium". 36.* Show that the production of a pseudoscalar neutral pion during a p-\-p collision is forbidden, if the neutral pion emerges in a P- state. Assume that the reaction takes place near the threshold so that the effective relative angular momentum in the final state is equal to zero. 37.* Which are the eigenf unctions and eigenvalues of the isotopic spin for a system of two nucleons? 38.* Show that d<3(p-\-d^*d-\-n-\- ic+) where the do are the differential cross-sections of the corresponding reactions taken at the same relative energy, angle of separation and relative orientation of the spins. 39.* Show that the differential cross-sections for the production of pions when accompanying the formation of a deuteron when two nucleons collide p-\-p-+d-j-x+ ■ (1) and n-j-p-+d-j-n (2) are connected by the relation d° (p + p-+d + 7i+) da (n + p -> a -f jcO) ~ z - 40.* Show that the differential cross-section for the processes n-\-p-*p-\~p + Tt- (1) and n-\-p-+n-\-n-\-T:+, (2) taken at the same relative energies, angles of separation and rela- tive orientation of the spins are the same. ANSWERS AND SOLUTIONS ANSWERS AND SOLUTIONS 1. ONE -DIMENSIONAL MOTION; ENERGY SPECTRUM AND WAVE FUNCTIONS E » = 2^ n ' +» <*> = V « sin T *' 3. a(p)|«= 4l,, * a fi //7 2 a3 (£- ")' cos 2 ^? for odd « . ^ pa sin 2/> f° r even n. 4. fi2 rf*|, ***+«?- v.) ♦=<> <*< 0) ' If we introduce the following notation y __YWW^EE. »_LzM „ _ Y2v-(Vi-E) *i— fi - x- - , x 2 g , we find that the general solution in each of the three regions has the following form: <|t = A x e- % ^ x -\- B^ x (x < 0), ty = Ae-** -\- Be** (0 < x < a), ty = A 2 e ~ x » x -+- B z e x * v (x > a) . Let us consider the discrete spectrum E < V 2 . y x and * 2 are then real. If we put in the interval < x < a x = /&, where k is real, we can write <|» = sin (ifejc-f-S) (0 < x < c). Since the wave function must be bounded, we have A t = 0, £ 2 = 0. 60 ANSWERS AND SOLUTIONS d<b The condition that <J> and ~ must be continuous can be expressed by requiring the logarithmic derivative t-^ to be continuous: x x = &ctg8, — x 2 = k ctg (ka -\- 8). Expressing x x and x 2 in terms of k, we can rewrite these last conditions as follows: Fig. 20 Since the cotangent is periodic with period it, the values of 8 and of ka-{-b can be written as follows: * . hk* , o = arc sin , -\- ti{R, YWt ka-\-h = — arc sin hk YW* ■f-ztgic. where the value of arcsin lies between and V^tf. Eliminating 8 we find a transcendental equation to determine the discrete energy levels hk . hk ka — ni: — arc sin YWt ■ arc sin Y^v* ' k =YM >0 . The values of k satisfying this equation must be obtained graphically, by finding the intersection of the line y = ak and the curve y = nit — arc sin hk YWx arc sin hk V>K 2 (see fig. 20) Let us consider a symmetrical potential well v l =V 2 =V. It is easily seen that in this case there is always, for all values of V and a, at least one level. If ^ fl < i, one can without difficulty find the value of the only discrete energy level. Expanding •jr — 2 arc sin in/! * in a series, we find E = V- pa* V 2 . The number of levels for arbitrary values of V and a will be equal to A/, where N follows from the relation 1^ nh ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 61 5. b) The eigenvalues of the operator //' are always positive. If ipo is the wave function corresponding to the state of lowest energy, we have therefore (Q + tP)% = or (JL + q^o. Hence we find that I 8a=— - % = c e 2 , eo __ In this way we find that the energy of the oscillator is equal to s n = n-j~~ («>0), and the corresponding wave function has the form The normal! sing constant A n follows from the condition f <b n (Q\ dQ = 1 -OQ For the ground state the normalising constant A is equal to _L i _i?l * 7 We can express as follows the wave function of the «-th state in terms of the wave function of the (n - i)th state, where c n follows from the condition Replacing P by — ^ and integrating by parts we find c lf%-i(P 2 -\-Q 2 -hD%- 1 dQ = cl-2n=l, so that c n = -_L= and = c n • Cn . x ... Cl (Q—^) n % = A n (q-^JV Finally we get 1 1 / d \ n - — 62 ANSWERS AND SOLUTIONS The polynomial of degree n is called a Hermite-Chebyshev polynomial. c) a a+ — a v a = 1 ; <!»„ = y= (<* + ) M 'V- d > (p-f/Q):_ 1 -[(^-^-J*^ 2,t - The wave functions <j> n we have taken to be real so that the matrix elements of Q and *P = -^ are also real, or, going back to the original variables, we have (*c 1 =(*)r 1 =v / ll' 7. The probability to be found is given by oo j" e~ y 'dy w = '■ :0.16. CO 8 The wave function must tend to zero for x = 0. For x > it satisfies the differential equation of an ordinary oscillator. One sees easily that the wave functions of the oscillator for odd values of n = 2k+ 1 tend to zero at x = and are in the region x^O. solutions of the problem under consideration. Hence £ k= =A«>(2ife+|) (* = 0. 1.2. ...) 9. ONE-DIMENSIONAL MOTION. ENFRCY SPECTRUM. WAVE FUNCTIONS 63 10. An investigation of the behaviour of the solutions of the SchrOdinger equation as *-oo shows that ip behaves asymptotically as exp (-fc) where 4 is a new independent variable e =Vp* 2 . ha As x-+Q <j> is proportional to W 2 , with We make the substitution and find for «(f) the following equation: v 2 / L2^4 2fiy2in7o J ( ; Equation (1) is the equation for the confluent hypergeometric function and its general solution is of the form *<0 = ^(«.v+y.5)-|-c^(a-, + | f |. — v,6).^' : \ where we have written a for the expression within square brackets in equation (1). From the requirement that tf>(0) must be finite, it follows that c 2 = 0. Apart from this we must require that the wave function should de- crease as *->oo, that is, that the function «ft) reduces to a poly- nomial. This can be attained by putting a equal to -»(n = 0, 1, 2, . . .) so that we find the energy levels The energy spectrum is .thus the same as for an oscillator of angular frequency <,, = j/jg provided the zero of the energy scale is suit- ably chosen. It is of interest to note that the zero-point energy of a particle in the potential V {±-±) 2 is always larger than the zero- point energy of the corresponding oscillator. The wave functions 64 ANSWERS AND SOLUTIONS are of the form V« — c n x e ' \ "• ^ ' 2 r A% 3 / where -, ... _ 1 (i/~ Sv ' v f i ~ ' \ 1 j-- l). while the constant c n must be found from the normalisation condition. 11. In the Schrodinger equation we make the substitution The equation for a takes the following form where •a a a rfjr~ a» v ' — K 2/j2 (we consider the discrete spectrum, E < 0). If we introduce a new independent variable z = — sh 2 — , a then the equation for u reduces to the hypergeometric equation 'O-^S+It-^-^'Is-^' 580 ' (1) The parameters a, % T , which occur in the general form of the hypergeometric equation have in our case the following values: 7 = y . a = v. — X, £l =,= — /. — X. Two solutions of equation (1) which lead respectively to the even and the odd wave functions ip are of the form u l== F(— >. + •/, — >. — x, y ;z )' (2) ^/iW-X-f-x + l, „>._., + -l, §;*). (3) These solutions lead to the bounded values of the wave functions at x — (z = 0). ONE- DIMENSIONAL MOTION. FN KHGY SPECTRUM. WAVE FUNCTIONS 65 In order that the wave function ' * = (ch£) '"a tends to zero as x-^-^ioo (z -► — oo), the hypergeometric functions in equations (2) and (3) must be reduced to a polynomial. Th'^ con- dition means, for instance, for u, that either X — v., or X-j-x must be a non -negative integer. The second case can, however, be dis- carded hince in that case the wave function increases exponentially as x-*+co We find thus X — x = k (k = 0, 1,2, . . .) and for the energy levels F — — l% \ l l/ HlxV ^ l"T 9b l V We find similarly for expression (3) that the condition that the wave function be finite as x->±oo is satisfied provided X — x — ~ = l (/ = 0, 1,2, ...) and hence Z' — S&btV -/;-+! ~(2/+ D- T J • Combining these expressions we find (n=--0, 1, 2, ...). The number of discrete levels is equal to the largest integer satis- fying the inequality A^/Sgt+f We note that the energy spectrum which we have obtained coincides for a suitable choice of the parameters with the spectrum of a Morse potential (see problem 11, section 8). 12. We make in the wave equation -£&-(*-".«< *)* = <> the following substitution ^ = (sin Joe) a. 66 ANSWERS AND SOLUTIONS Putting we get the following equation for a -7-5 — 4 — X ctg — - — — 5- (v 2 — X 2 ) u = 0. By introducing as the independent variable 2 = C0S 2 — a the last equation is reduced to the hyper geometric equation Comparing this equation with the general form of the hypergeometric equation we find the parameters: -f==— , a = — n — X, p = v — X. Equation (1) has two solutions. One of these solutions is different from zero and finite at 2 = (which corresponds to x = j) , «i = / ? (— v — X, , — X, l;z). The other solution „ 2== -^(_v-x-hi, v-x+^. |-; *) tends to zero at z = (* = y ) . To determine the behaviour of the solutions at 2=1 (which corresponds to jt = 0, or x = a), we use the relation F(a, p, 7; 2) = (1 — z)~ a p(a, T — |3, t: jzrf)- We find for "1 and u 2 Bl = (l_z)» + ^(-v — X, -v+X+i, ^-; -L-), (2 ) v+x-i B 2 =r/Z(l— «) 2 X XF (_ V _X4-^, —v + X+l, |; ^-f). (3) In order that the wave function $ tends to zero at x = and jc==o it is necessary that the power series in — ^r is a polynomial. ONE-DIMENSION AI. MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 67 The hypergeometric series in expression (2) for u t breaks off if either v-f/. or v — \ — _L is a non -negative integer. However, the condition ip - at x = and x = a is satisfied only in the second case, v — \ — l = k (* = 0, 1, ...)• The energy levels are now given by the equation E, = [( 2A> + 1)2+4(2*+ l)X-2X]|g. (4) A similar discussion of equation (3) shows that the energy levels satisfy the condition v — X = / (/=1, 2, 3, ...). (5) We cin combine expressions (4) and (5) for the energy levels as follows, f fl = (»*+4«X — 2X)^ (*=1.2,3, ...)• For odd values of n we get the corresponding wave functions t.-«.(-n2f ^(-$-2*. », | ; cos ^), while even n corresponds to ♦.= <„(- ?)"*««? Xf(+a + }, f-f+ f ; a*«L). The normalised wave function of the ground state is given by the equation 6 = Let us consider the limiting case where V -+Q. The problem goes then over into the problem of a particle in a potential well (see problem 1, section 1). The quantity X tends to zero and, as we should have expected, we find for the energy levels E = r ™n* In the opposite case where 1>1 we find for the low lying levels (»<C*) £„=fi«>(rt-h-i) (»=1,2, ...), 68 ANSWERS AND SOLUTIONS where _ rW : a r jj. This result could also have been obtained by expanding the potential energy near the point x =^ and retaining only the quadratic terms. 13. In the case under consideration there is only a continuous spectrum and the eigenf unctions are non -degenerate. Let us in the Schrodinger equation 2\x dx* go over from the coordinate to the momentum representation; we get £a{p)—Ea{p)=thP±a<j». The solution of this equation corresponding to an eigenvalue E, a E {p) = ce~ **{**■ Ep ' is a wave function in the momentum repre- sentation. We normalise the function a (p) to 8(£ — E'), fa* E (p)a E , (p)dp = MF-t?). or ip (B-E>) cc*fe hF J 'dp^cc*2T:hn(E — E'), whence l Y2xhP The wave function in coordinate representation is -I oo 3 <-o dt (x) =- = e A du = —7= cos (■- — uq) du, YV ' 2n VP .' n YP J V3 / — oo This integral can be expressed in terms of the Airy function <fr(q): oo * (?) = 7^ J cos (£ + uq) du, * (x) = -A_ * ( __ , } . 14. The Hamiltonian operator has the following form for the given potential ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 69 Since s_ e d P a (p) = a (p -f- bti), the Schrodinger equation can be written as a finite difference equa- tion ±p*a(p) + ±V a(p-t-bh)-\-^Vo"(P — bh) = Ea(p). | *« (A) + "f V„a (k + ^) = ^ (*) — oo (+°° 2r.nix \ — oo 16. The wave function in the region of the well (0 < x < a) is of the form -i = Cl e ix > x -+- c 2 e~ ix ' x , x x = T 2 * E , and in the region of the barrier — b < x < $ = * 3 e'«* + c 4 e- '■**, x 2 = V2t*(g-V ) Since 6(x) = const • $(*+/) (the constant has an absolute magni- tude of unity, /= a-\- b) in the region of the next barrier (a < .v < < a _^_ b ) we have 6 = e ikl {c 3 e u ^ x -^ -f- c 4 g-» x * (■*—>)). The requirement that the wave function and its first derivatives are continuous at x = Q and x — a leads to the following four equations: c^ a -+- c 2 e-^ a = e m {c z e~ ,x » 6 -+- c i e iA - h ), x^q — c 2 ) = x 2 (c 3 — c 4 ), Xi (c^Xo _ c 2 e-«'*.a) = y . 2 (c 3 e-**.6 — c^'"** 6 ) e iW . (1) This set of equations has a non -trivial solution only if A + A ■ ■ a cos kl = cos x,c • cos y.Jb k- sin v n a • sin v. 2 0. We investigate two cases: a) £ < V , x 2 is purely imaginary. Introducing the notation x 2 = r/, we can write equation (i) in the form 2 2 cos kl = cos x t c • ch %b -\ 2~r~ sin * ia ' Sh %b ' ^ 70 ANSWERS AND SOLUTIONS The allowed energy bands are thus determined from the relation 2 2 — 1 ^ cos x t a • ch -/.b -\ j — - sin % t a • sh *b <; 1 . To discuss the general regularity in the distribution of the allowed bands we consider the limiting case ^Ms.b<^\ t a^>b, E<CV . We introduce i^ fl ^_ T . Equation (2) is thus approximately of the form , sin A.,a , cos ka = i ! — h cos x,a. In fig. 21 we have plotted the function U £H12l£-|_cos x x o) as a function of x x a. The allowed energy bands are indicated on the x t a-axis by heavy lines. ,,^-^s'mx,afCOSx,a K=XS W A l\ 3tc / x,a in \l 17 vn H£ Fig. 21 On the right of each point x x c = n- there is a forbidden energy band. From the figure it is clear tnat the forbidden energy bands get narrower with increasing number of the band. One can easily estimate the width of the forbidden bands. The left hand side of expression (3) takes on the values (— l) n , when cos(x 1 a — <s) = ( — l) n cosc5, teres = ^-_ which is possible for v^a^n- and for x,a = «--{- 2 ? . It follows thus that the width of the forbidden energy bands is equal to 2?. For large values of n 23 ONE-DIMENSIONAL MOTION. ENERGY SPECTRUV. WAVE FUNCTIONS 71 b) E > V . In that case the energy bands are determined from the relation -/? -f- % 2 — 1 < cos y.fl • cosv. 2 b k sin x,a • sin v. 2 b < 4 } • 17. The energy levels E n are determined from the quantum condition -«-2 jprfx = ic(n+y)fi, (1) where P = /*p5| and where x x and a: 2 are the turning points determined by the con- dition #= (£„ < for the case under consideration of a discrete spectrum). To evaluate the integral '< £ >=J/ 2 f+i|) dx we differentiate both sides with respect to E. As far as the differ- entiation of the upper and lower limit of the integral is concerned, that contribution vanishes since the expression under the square root is equal to zero in the points x t and x 2 . We get thus di_ = r dx dE ~~ M r 7 Vo - ^ 2m- / £ + , Ch2- Introducing the variable sh — = z the last integral reduces to the following form dl |* dz P-Q We find thus /(£) = — V^— 2ixa 2 £ « + C. The constant C is determined from the condition that for E = — V the domain of integration reduces to a point so that f(-V ) = 0, 72 ANSWERS AND SOLUTIONS and hence 1(E) = 1^2^(1^^ — V—E) it. The energy levels are thus in the semi -classical approximation determined by the following equation F — — wW^-i'+r)]' <» = 0.l.2....).(2) The number of levels N is equal to f ° - a. We note that the evaluation of the energy levels by means of the quantum condition (1) is legitimate if the number of levels is large, that is, if 2jXfl2^ If that condition is fulfilled, expression (2) for the energy levels coincides with the exact equation for E n obtained in problem 11, section 1. 18. a) £„=(/i+-i-U«o (» = 0, 1, 2, ...). b, ^[V^+KI)]'-". (n = Q, 1, 2, ...)• 19. The average value of the kinetic energy in the stationary state ty n (we assume the wave function to be real) is given by the equation 2f» J V n dx* ax — 2ji J\dx) aX ' In the semi-classical approximation the wave function has in the classical accessible region (a < x < b) the form ( x \\ pdx ~^)> P-VWi^=v], so that dx ^\?\><*-^-\%%A^U* If we substitute this expression into the integral which determines T the limits of integration must be the limits of the classical acces- sible region, since outside that region <|» n decreases exponentially. Taking for the square of the fast oscillating trigonometric functions their average value V 2 and neglecting the integral containing the ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 73 oscillating product -'■(tJ""-t)-»'(t/^-t)-t-(t/'»"- f). we get T = -^\[ p + W^] dx - The condition of applicability of the semi-classical approximation I d% I h I dp I „ , I "57 1 = ^ 1 5J | <C 1 means that the second term under the integral sign is small compared to the first term so that we find by using the quantum condition a The constant A n follows from the normalisation condition a \ a I a On the other hand, differentiating the quantum condition O b J pdx= JY2 v .(E n — V )dx =,/i( n -f') a a ' ' with respect to n we get dn J y2|*(£ B -lO r </„ J p O a so that "t/j rf/2 The expression for the average kinetic energy after substitution of the last expression takes the form Y _ l dE n 2 rf« («+±)- 20. a) f I= ^ So ,(„ + ^. 74 ANSWERS AND SOLUTIONS 21. From the virial theorem it follows that if = W, so that o ' .j - Substituting into that relation the average value of the kinetic energy (see preceding problem) we get the equation e = ^"(»+t). 2v rfn whose solution is of the form £ = const ( /i + "2") 2+V 22. We start from the quantisation rule |V2ii[E — V(x)]dx = *b(n-{-^), X, which determines the spectrum, or more precisely n{E), if the potential energy V (x) is given. Inasfar as V(x) is an even function we have * 2 \V2p[E — V(x)] dx = ::fi (n + 2 ) ' (1) where x 2 = — x l = c, £ = V (a). The problem is thus reduced to finding the solution of the inte- gral equation (1) which is of the form * v dE *°° = wJ-« yv — e F u dn where x(V) is the inverse function of V (x) and d A is considered to be a function of E, while E is the zero of the energy scale. 23.* The Hamiltonian of a free plane rotator reduces to the Ml , ,» operator of the kinetic energy which has the form ~- , where M z is the operator of the component of the angular momentum of the rotator along the axis of rotation, the z -axis. * This problem is closely related to the problem of classical mechanics of finding the potential energy if the vibrational frequency is given as a function of the particle energy. ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 75 The Schrodinger equation for the stationary states of the rotator takes the form (we replace M t by -j-g- where cp is the angle of rotation of the rotator around the z -axis): 2/ d^ ^' The general solution of this equation which satisfies the condi- tion of uniqueness (which in this case is the requirement of periodi- city in cp ) is obtained by putting -i/~^£ equal to an integer m (m = 0, ±1, ±2, . . . . ) and has the form <W» (?) = c i s ' in m ? 4- C 2 cos m ? » where e t and c a are arbitrary constants, and the corresponding energy levels are according to the considerations of a moment ago equal to E„ = Writ* 2/ As could be expected, each level for m ¥" is twofold degen- erate with respect to the direction of rotation. The normalised wave function of the stationary state corresponding at the same time to a well-defined value of the angular momentum (M z = mh) is clearly equal to 24.* The potential energy of the particle under consideration is of the form f oo, £<;o, where the direction -g is, as usual, taken along the positive 2-axis and where z = is the reflecting plane. This potential energy and also the energy levels and wave functions which we shall obtain are depicted in fig. 22. We restrict ourselves to an investigation of the motion of the particle in the z- direction since the motion in the jc^-plane is free and is of no interest at the moment. The time -independent SchrSdinger equa- tion for the particle wave function ty(z) in the region z>0 is then: Fig. 22 76 ANSWERS AND SOLUTIONS -2^^ + m ^ = ^« (1) where E is the particle energy. This equation must clearly be solved with the boundary conditions ^->0 as z-^oo (2) and ^ = for z = 0. (3) To simplify equation (1) we introduce a new independent variable £ which is connected with z by the relation 2^ z _2mE ==cr , t (4) where c is for the moment an arbitrary constant. Substituting this variable into equation (1) transforms it to rft 2 C \2nfig) 'VW = 0- V It is now clear that if we take c = f "If -) ' , that is, write equa- tion (4) in the form we are led to the very simple form <T(C) — 0KC) = 0. The solution of this equation which is finite for all values of £, that is, also for all values of z, is oo q = A$Q==A-^jcos(u>:-t-luAdu, (5) where $(£) is the Airy function and A a normalisation factor to be determined in a moment. To begin with it is clear that the energy levels of a particle which is in the potential well shown in fig. 22 form a discrete spec- trum and that their number is infinite. To find these levels we use the boundary condition (3) which, by means of equations (5) and (4'X can be written in the form $ LfLn. (3') ,)-. x m '"g lj n u Putting the argument of equation (3') equal to the roots of the Airy function which we shall denote by — a n (n = 1, 2, ...) ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 77 °<«i<a 2 <...<a n <..., we find for the solutions E n of this transcendental equation Hence the required energy spectrum of the particle has the form £« = (-^-)'-a n («=!, 2, ...). (6) In particular, the energy of the ground state (n= 1. a, ^2,34) is equal to *,» 2 .84(i£i)\ (6.) As it should be for the case of a one -dimensional bounded motion, the energy levels we have found are non -degenerate. The wave functions of the corresponding stationary states are according to equations (5) (4'), and (6) of the form +.W = A.*(C)«A.*(|-« Ii ). (7) where a denotes a characteristic length n ( m \ ,/3 The normalising factor A depends on n and a. This function is oscillating in the classical region, and the number of nodes of the function ip n is equal to n (including the node at the edge of the region for * = 0), while the distance between consecutive nodes which is equal to * uX = - F - * ft ==-, (9) Y'2m[E — U(z)\ decreases towards * = (that is, with increasing kinetic energy see fig. 22). * y It is clear that the quantity given by equation (9) has this meaning only for n » 1. In this case the bar denotes an averaging over a section \z which con- tains e large number of wave lengths 2nX, but which is small compared to the dis- tance over which U (z) varies appreciably. 78 ANSWERS AND SOLUTIONS The constant factor in equation (7) is determined from the nor- malisation of $. If we normalise with respect to £ this constant is clearly equal to * n ~ / — * /7 [*(C)]*rfC ( 10 > and if we normalise with respect to z, we have [ see equations (4') and (8)] j j A. /7 r -a We investigate now the asymptotic behaviour of the wave func- tions ij) n {z) at sufficiently large distances from the classical turning point ** ,c1 = JLl MJ\ and also the form of the energy spectrum E n for n^> I, that is, in the semi-classical case. According to equations (6) and (8) the lengths z„ and a are con- nected by the relation We use equations (8), (11) and (11') to write equation (4 T ) in the form _ z d z ^ a a n and consider asymptotic expressions for the functions ty n {z). 1) The classically inaccsssible region and values of z not too near zf : . z-zi^>a. i.e. C>1 *** * As the Airy function is real for real values of its argument, we have |*(0l a = [^(01 a . ** z^ 1 is cleariy the greatest height which a (classical) particle of weight mg and given energy E n can attain. *** It can easily be shown that the inequality ( £( » 1 is equivalent to — <^T 1 by virtue of equations (9) and (4' ). ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 79 Using the asymptotic expression for the Airy function 2C'' $>(r)^_U-^ V ' v ' or'/i W6find /. .CINV. ^>~H^r " (12) (exponential decrease of the probability |^ n | 9 in the classically in- accessible region). 2) The region of the classical motion and values of z not too near zf „, , 0<z<zf amd\z — zn\^>a, that is, — o B <C<0and|C|^>l. It is clear that to satisfy the second condition at least the follow- ing inequality must obtain: a n ^> 1. As we shall see below, a n is proportional to n 3 which means that the interval of £ -values con- sidered here exists only for very highly excited (semi -classical) energy levels, that is, for n^>l. The required asymptotic expansion of $(f ) has the form *ro~fiV sln (f |C| ''"+7)' (13 > *»<*>~H^V sin h(hr~) + t]- (i4, The asymptotic expressions (12) and (14) give us in this way the semi- classical wave functions. By means of (14) one can find explicit ex- pressions for the roots of the Airy function C n ss — a n , and hence also for the energy levels (6) in the semi -classical case. To do this we must clearly equate the argument of the sine to an integral multiple of 7T: TlClN-HOi+l). (» = 0, 1,2,...), so that «»=IU = [|-(/»r + ^-)]\ (15) and [see equation (6)] E n = I <?«*mfb*)'''(n + ff. (16) 80 ANSWERS AND SOLUTIONS The last result can, of course, also be obtained without knowing the wave functions by using the Bohr -Sommerf eld quantisation rule (see problem 27, section 1). Simple expressions may also be obtained for the normalisation constant A n in the semi-classical case. Indeed, since for n ^> 1 the wave function (7) oscillates rapidly in the classical region < z < zt [ see equation (14)] and outside this region decreases exponentially, we can in the integral of equa- tion (10) replace sin 2 by its average value V 2 and limit the integra- tion over £ by t= 0. Taking into account that «>lwe get thus: and [see equations (10) and (10')] : An ^ l (JL\\ (17) " Ya \3nnJ Let us now write down an expression for the average distance ^X [see equation (9)] between consecutive nodes of the semi- classical wave function. Using equations (4') and (8), we find: * VWn /It— r Since in the semi -classical region |CJ >1, we have *% <C a. Near the reflecting plane, that is, for L <C a n> this "semi- wavelength" is a minimum and as it is equal to 27l2 v, a (**)-~(— ) -&' (18) Finally, the distance between two successive levels AE n follows from equation (16) which leads to ai p A£ " 2 An AlnE n =- T -=3— , and by putting Arc = 1 to: A£ =11". (19) * The same result could have been obtained by using equation (13) and the obvious relation taking into account the fact that n% == ±z = a -±+. ■ ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 81 Let us now show the limiting transition to classical mechanics. This limit can be obtained by averaging the quantum mechanical pro- bability density I^J*)!* over an interval Az which is small com- pared to the dimensions of the classical region of motion, but contains a large number of wavelengths, so that it satisfies the inequality: 2*X <C Az <C z f. According to equations (14) and (17) we have: , t . W |.=i(^)' 4 1 /=z^: sll ,.r£/iiz<!if + -]. a \3-nJ y \z-z$\ [3\ a / ^ 4 | Averaging this expression over an interval Az we can replace the square of the sine, which is oscillating many times in that inter- val, by its average value 1, which leads after multiplication by Az to: 2 R^iyp Az = f- 2 -) Va _J Az. (20) K.lr.nJ 2Va\z — zf}\ V ' In the classical limit (fi-*0, /1-+00) this probability must go over into T7T ( sin ce it is normalised) which is the ratio of the period spent in the interval Az to the semi -period of the classical motion. This means that in the limit we must have |*(z)|*Az=-j At J T Eliminating from equation (20) aW* ~ (th)''' by using equation (x6) and omitting everywhere the index n we get £=-17 /IT < 2i > i 2 where H - \z — z c |. This ratio must clearly be equal to the classi- cal velocity of the particle ItII^ci^v'"^. Since F max iitfj 2 ~ J v - J YWl ^* '"^' < 22 > the equality — = y~2gH is, indeed, satisfied. In conclusion it is useful to estimate the order of magnitude of the quantities which we meet with when we apply the' results obtained 82 ANSWERS AND SOLUTIONS here to macroscopic particles, for instance, a ball of mass m = 1 g, freely falling from a height H= 100cm in the earth's gravitational field (£« 10 3 cm/sec ~ 2 ) on a solid slab. The characteristic "zero point" energy e^ (rngVpft [ see equa- tion (6')] is of the order of magnitude •(I • lQ fi 10- 5, ) 1/3 = 10- 16 erg, and the energy of the ball E = nigli^ 1 • 10» • 100 = 10* erg, so that the number of the state (the number of nodes of the wave function) n and the distance between adjacent energy levels AE n are given by [cf. equations (16) and (19)] : T) 10 3a , 10' erg. to: The characteristic length fl ~(i) * [see equation (8)] is equal ir!;y /3 =io- 19 1 • 10 3 / cm and the (minimum) wavelength of the ball is equal to %■• U(x) as it should be, since a • n~H . The magnitude of these quantities gives us an idea with what enormous precision the motion of macroscopic bodies follows the laws of classical mechanics. 25.* To avoid computational difficulties, we consider first of all a symmetric well of very great, but finite depth U (see fig. 23) and later on take the limit U -+oo . We evaluate the average force / exerted, for instance, on the right hand wall of the well by the particle. The operator of the force exerted on the particle by the external field (the wall of the v well) is equal to f. dU {x) dx where U(x) is the 23 operator of the potential energy of the particle. According to Newton's third law. the force exerted by the particle on the wall is equal, but opposite in sign, to ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 83 that force, or /<*)= dx dU(x). (1) dx ' and its quantum mechanical average (which we require) is equal to: CO 7= f Hx)mx)\*dx, (2) —oo where ty(x) is the (real) wave function and f(x) is taken for the (right hand) wall under consideration. Since U(x) is everywhere constant except at x = a where U(x) jumps suddenly by an amount U Q , equation (1) may be written in the form: f{x) = U Q l{x — a\ (3) .where 6 is the delta-function * . Substituting expression (3) into equation (2), we get: 7=I/ l<Kfl)l 9 . (4) To find ip(a) we consider the solution of the well-known problem of a rectangular symmetrical well of finite depth. The wave function is of the form (see problem 4, section 1, with V l == V 2 = U and m instead of n for the mass of the particle): <}»(x) = c 1 e xx (*<0),1 where (5) where * = /£.(£/„-£). J ^(x) = csw(kx f 3) (0<x<a), sin (ka -[-&) = — sin o = (6) Y'2mU ' i i {x) = c 2 e~* je (■*><*)• (5 T ) * Indeed, from equation (1) we get (e -» 0): a+a j f{x) dx -^U(a + e) — U (a — e) = £/ ; the same follows clearly also from equation (3). 84 ANSWERS AND SOLUTIONS According to equation (6) the value of #(a) in which we are interested is equal to: * tik <b (a) — c sin (ka -\- o) = — c Y'2mU ' (7) In accordance with equation (2) the wave function must be norma- lised. From this requirement, we can immediately determine the coefficient c if we take into account the fact that the normalising oo integral f \ty(x)\*dx = 1 is appreciable only in the interval 0<x<a. — oo Indeed, outside that interval the wave function decreases exponen- tially according to equations (5) and (5') over a very small distance of tne order of magnitude J_~__A__ (we remind ourselves that we are interested in the limit U -+oo; for the same reason we can neglect E compared to U Q ) so mat the contribution of the regions *<0 and x>a to the total probability f Wdx is negligibly small. — oo We have thus oo a a 1 = J [<]f(x)\*dx^i J \<!{(x)]*dx = c* I fAn*(kx-\-$)dx = -oo = c 2 -H" — or s i n ka • cos (ka -\- 23) . /8) Let us now take the limit (/„->■ oo. According to equation (6) we have for U Q -+oo : 8-> wr, (/ta + o) -> HjH and consequently fea -> (n, — «) n where »■ and n { are integers, so that the last term in equation (8) tends to zero. Indeed, 7TT- sin ka cos (ka -f- 2tt/i) = — sin 2ka -> jt sin (2- (n 1 — n)] = 0. We have thus for a well with U = oo: ]/ f (9) Using equations (9), (7), (6), and (4), we find finally / = -f- (10) As U Q has disappeared from the result, this is also valid for the limiting case where U — ► oo . The force / is obtained clearly with the correct sign (/> o on the right hand wall of the well). (The average force acting on both ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 85 walls is clearly equal to zero. Indeed, the operator of this total force is U Q {*{x — a) — 6(jc)>. so that we find, taking the symmetry of the problem into account: /==£A>{[<M«)P-['MO)]2} = o. This result is correct for finite motion in any arbitrary form of the potential energy. ) We note that the result (10) which does not contain h explicitly retains its form exactly in classical mechanics. Indeed, in that case / c i (time average! ) is equal to the product of the momentum imparted to the wall at each collision (2mv) and the number of col- lisions with the wall considered per unit time JL (v is the par- ticle velocity), or 7 o v 2E 26.* To find the probability distribution we want, we expand the wave function of the (non -stationary) state under consideration in terms of the wave functions ipn of the stationary states of the particle in the well, where •H*) = 2cA>(*). (1) ^W = F 7 sitl T x (»=1, 2, 3, ...). ( 2 ) The probability to find the particle in the n-th state is equal to: w n = c, f ty(x) ¥„(.*) <t* (3) and fX-Jl+WN*. To obtain the normalised probabilities w n we normalise the original wave function. To do this we must take = 30a- 5 . Substituting ip and ip n into equa- A* = ( x*(a — xfdx tion (3), we have: w n 30 # 2_ a 5 a /. . T.nx , x(a — x) sin ax 86 ANSWERS AND SOLUTIONS Evaluating the integral we have finally: ™„=|^ 8 [l-(-l) M l a . (4) The probability «V, = c* which we have obtained is different from zero only for /»= 1, 3, 5, . . . These values of n correspond according to equation (2) to even (with respect to the middle of the well * = £ ) functions $ n (x). Thus in the superposition (1) only even states are represented (as should be the case, since the origi- nal wave function is. even with respect to x = -^ ). Let us verify the normalisation of the probabilities w n . We use the equation yi i _ (2 9Bt -i)« Bw ,o , 24 ( 2k-lf m ~~ 2.(2/»)l ! 2?wI ' < 5) where B 2m are the so-called Bernoulli numbers, and we have (in our case 2m = 6, B 2m = B 6 — -^ )'• OQ CO OO 240 rtQ Vl 1 960 (2* — l) 8 n=l n = l. 3,5, ... 7v = l 960 (26— l)7l6 J_ = "^ u« ' 2-61 '42 as was to be proved. According to equation (4) the probability to find the particle in an even state corresponding to a quantum number n decreases fast with increasing n. The probability to find it in the ground state in = 1) is equal to: 94ft «,==?• 2«« 0,999. so that the total probability to find the particle in all the excited state (» = 3, 5, . . .) is equal to 0.001. This is connected with the fact that the original wave function is everywhere very close to the wave func tion of the ground state of the particle in the well One can thus say that the state under consideration is "almost sta- tionary' ' . * * This nomenclature is only used for this particular problem and should not be confused with the conventional term "quasi-stationary state", which has a com- pletely different meaning. ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 87 The energy levels of the particle in the well potential considered are given by the equation: E » = -2f^ n3 (*=1.2,3,...). (6) The average energy of the particle in the non-stationary state considered is, according to equations (4) and (6), equal to oo oo e=\ew =^- 240 - 22 y -. n»i n=i,3, 5, ... Using equation (5), we find (2m = 4, £ 4 = — ^) : 2 /"^ = 96* so that ^ n=l, 3, 5 £=,*£. J^l.014^. (7) 2jj.a 2 n 2 * where E x is the ground state energy. The same result (7) is, finally, obtained by taking the average of the Hamiltonian H= f-\-U over the (normalised) wave function, 5 f .•£. ^ *■ f , ^ a f> 2a5 42 5fi2 E=J^rf* = — -^-Idx^-^A***^. {V) o The average of the square of the energy of the particle in the state under consideration is equal to: oo oo *=2**-(#J-S S £-*>(£?• < 8 > !* = 1 W = l, 3, 5, ... According to equations (7') and (8) the energy dispersion is equal to A P — 1/ <P P\2 — 1/ P2 fp'te — l/"* LE = V{E-Ef = |/"^-(£> = ^5 -^- < 9 > which is comparable with EmE v This relatively appreciable value of the energy dispersion is clearly explained by the appreciable contribution of the excited state to the quantity E* (by virtue of the fact that E\ ~ n ). 27.* The potential energy of the particle is of the form (see also problem 24, section 1): f oo («<0). \mgz (z>0). One of the turning points is z = and the other one follows from the equation U(z) = mgz = E and has thus the coordinate: 88 ANSWERS AND SOLUTIONS E me; As far as the quantisation rule is concerned, its right hand side must be slightly modified compared with the usual expression n(n-\- ^-). This is connected with the fact that in the usual derivation of the Bohr- Sommerfeld quantisation rule one assumes that the potential energy near both classical turning points z = a and z = b can be expan- ded in a power series in (z — a) and (z — b) beginning with a linear term. As a consequence in the semi-classical expressions for the wave function, starting from both turning points, the phase starts at — : Fig. 24 and Yp p{z)dz-\-^ I («<*<£). (1) [b s-J ' «_ The requirement that these two expressions are the same in the interval a < z < b leads to the condition 6 ^- J pdz+~ = (n-\- l)u (n = 0, 1, 2, ...) (2) a (the total of the two phases in equations (1) is put equal to an integral multiple of n). In the case under consideration one of the turning points (z = a = = 0) corresponds to a "vertical potential wall" so that the first of the two phases of equation (1) does not start with a term -J (in contra- distinction to an "inclined" potential wall the function behaves here semi -classically right up to the turning point z = a itself where it must tend to zero, whence this result follow&J,' in the second of the phases of equation (1) the term j enters as usual. If we now add the two phases we get thus instead of equation (2) the condition b j J pdz-\-j = (n-j-\)K, a from which we obtain the quantisation rule we are looking for, i J /,</* = (» -|-|)ic (» = 0, 1. 2. ...)• (3) ONE-DIMENSIONAL MOTION. ENGERY SPECTRUM. WAVE FUNCTIONS 89 Substituting p=Y2m[E— U(z)] = V2m(E — mfrz), a = 0, b = -£- mg we get after elementary calculations £^E n = l(9^^A2) ,/ '(„-f-i) / ' l (4) which agrees exactly with the semi -classical result obtained in problem 24, section 1, as should be the case. If we borrow from equation (22) of that problem the value of the classical frequency of the bouncing ball 2* -./~m and evaluate from equation (4) the distance between successive levels (An = l) , we arrive at the relation **--# which expresses the approximate equidistance of the semi-classical energy levels. 28.* As the potential is periodic it follows that there must be solutions of the SchrOdinger equation for which <]) (x + a) = z^ (x). If \z\ =£ 1 the wave function increases unrestrictedly in one direc- tion or the other. The condition from which the allowed energy bands are determined is thus |2|=1 or z = eh. Let us now consider one period of the potential (see fig. 25). It can be divided into three regions: *, < x < x 2 , x 2 < x < * 3 , x 3 < x. In the first region the solution can be written in the form: Fig. 25 or, differently: ti=» Vp CO pdx Yp I P I 2 +7T k ! pdx -r) 90 ANSWERS AND SOLUTIONS where c ' i==c /(*-$), c' 2 = c^"^, « = ^ J Pdx. Using the boundary conditions of the semi-classical case we obtain a solution in the second region: -5" f I *>l*» l P * " 1 4 ^'"*' or where Id*. Using again the boundary conditions, we get for x > x s : to— ' v — ^ — »* -' /j « *■ • The condition that the solution Is quasi -periodic is of the form: -*(<-4--^)e T = *<2. Expressing c{ and c£ in terms of e x and c a we obtain the following set of equations: Cie u (g-P _f 4^) + /c 2 e- is («"* — 4*P) = 4*c t , Cl e*« (e-P — 4ef>) + /c 2 «-*« (e~P + 4<?P) = 4/zc a . The value of z is determined from the condition that this set can be solved (that is, by putting its determinant equal to zero): *a— 2*(«P + j*-P)cosa+l =0, , 12 = = ( e P+-i g -P)cosai:y r (^+i- e -P) 2 cos2a— 1. If the expression under the square root is positive, then z is real and different from unity. In the opposite case, z is complex and its modulus is equal to unity, ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 91 jri(fBas ^ + |,-p)cosa=t:/]/"l— ^? + I e -pJcos2a. |*i,.l 9 -(« p 4-T*" p ) tco * 9a+ + [l — («P + 4 e- ? )* cos 9 a] <- 1. Thus the equation which determines the allowed bands follows from the condition that the expression under the square root in equa- tion (1) is negative, (eP + l<H>) 2 cos2a<l. (2) The condition that the semi-classical approximation can be applied is that the expression f pdx is large compared to fi, that is, a » 1, > >► 1. If, in accordance with this, we neglect the term with the negative exponent we can write equation (2) in the form e 2Pcos a a<l or cos a a<g- 2 ^ (2') To a first approximation, we have cos 2 a = 0, a = nit -\- -~ or CDt This is the Bohr -Sommerf eld quantisation rule. From the solution of this equation we get a set of levels in a well which is a hole in a periodic potential with impenetrable walls: E , E lf E 2 , ...» E n , . . . We now expand the left hand side of equation (2 1 ) near the value a n corresponding to the »-th energy level cos 2 a « sin 2 a w (Aa n ) 2 «=; (Aa n ) 2 . From the definition of a we have A£ d T . A£ C dp . A£ ?dx__*ET &a = -FalJ P dx ~T) dk dx =T J 7~T7- 92 ANSWERS AND SOLUTIONS where T is the period of motion of the particle in the corresponding well -level. If we note that e-w is the probability of transmission through the barrier, W{E), we can write equation (2') in the form (AE n )3<±Jw(E n ), so that we get for the band-width: 2, ae„ = |V^(£„)<:k>. 29.* 1) The SchrOdinger equation in the momentum representa- tion is of the form: where OUh g-J = U(x) is the operator of the potential energy of the particle. We must look for a solution of this equation in the form: 0(p) = e-i 8{p \ (1) where we expand S(p) in powers of h: S(p)-S (p)-{-hS 1 (p)~h^S 2 (p)-i- ... (2) To find an expansion in powers of h of the expression which we obtained by letting the operator 0(lh ^-j act on the wave function C(p), we assume that the potential energy can be expanded in a power series in x. Thus, KO-t^Hl < 3) Let us consider the action of the different terms of this series on the function (1): op dp ' ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 93 In general, as can be easily shown by induction: {(v)"+ a (i:*)0(frvo«^ _ -r^((*)- + «-< 2F ii«(«)- +0W ,. (4) According to equations (3) and (4) we find The equation for the action function 5 (p) has thus the following form: Substituting now the expansion (2) for 5 and taking together terms of the same order in h we get the following equations to deter- mine S and S{. + "(£)-*• (5) ~df u \ dp )~ r 2 dp* u V dp )~ v - (6) Let x(p) denote the dependence of the coordinate on the momen- tum when the particle moves in the field U(x). We find then by sol- ving equation (5) with respect to Uln. , dp -dp~ = X (P). so that s o = / x (p) dp. From equation (6) we get: dp 2 dp* ytfdSny One sees easily that the right hand side is the derivative of -T'"Kt)l' 94 ANSWERS AND SOLUTIONS with respect to the momentum, so that *.—- H"'(f)l- The semi -classical wave function has thus finally the following form in the momentum representation: °<P>° A, '^t '"'^"' < 7 > V>'(f)l The meaning of the factor in the denominator is clear when we note that ^'(-^ll is the force, MJ- 1 , expressed in terms of the momentum p. This means, as should be the case in the semi- classical case, that the probability to find the momentum between p and p-\-dp is proportional to the corresponding interval of time dt: \0\*dp = ^dp==\c\*dt. 2) We now turn to the second part of the problem, that is, to show that the wave function Q (p) which we have found can be obtained as the Fourier transform of the semi -classical wave function in the coordinate representation. The semi-classical function in the x -representation has the form: Hx) vjp-m where p(x) is the momentum as function of the coordinate p(x)=> -V^lfi — U(x)\. The corresponding function in the momentum representation is* ^. . 1 f . . / v -«•**» Ci f dx T I p(x)dx-px\ /ov 0(p)c= dxb(x)e h «= r L— r . e h L J J. K°) Ky ' f2nh J TW /2wfl J V\p{x)\ Using the fact that the action S — j pdx is large compared to h we can evaluate the integral in equation (8) by expanding the exponent in a series near its maximum (saddle-point method). * T7e denote this function by G to distinguish it for the moment from the semi-classical function in the momentum representation which we found sub (1); our problem is to prove the identity of these two functions. ~dx ONE-DIMENSIONAL MOTION. ENERGY SPECTRUM. WAVE FUNCTIONS 95 To find the maximum we put the first derivative of the expres- sion <p (x) =3 J p (x) dx — px equal to zero: *=P(x)— p = V^[E — U(x)l—p = o. (9) The solution of this equation is x(p), that is, the coordinate as function of the momentum in the field U, or, the function which we considered sub (1). For the second derivative we find: d P , ^ dx* V^iE-U) ~ ? p(x) ' To find the value of <p" at the maximum we must take x (p) for x, and we thus find p[x(p)) = />, so that 'max r The expansion of y(x) near its maximum is thus of the form: <?(*) = J Vty(E — U)dx — px(p) + v >-4L—[ X — x(p)]*. If we substitute this expression into 6 (p) and take out from under the integral sign the slowly varying factor . * . at the maximum x = x (p) we get: p <*>{p) dp CB (j>) 4 [j^ix^-^lete-poJ^)] ri^~\p~\ f «(J»). ft 1 t II VlHE-U)dx-px (p)\ r | a | We prove now that the expression within square brackets may be written as —J x(p)d To do this we take its derivative with respect to p: <B(p) 7p-{\V^[E-U (x)) dx - px (p)) =%LY2 V .{E — U [x(p)]}-p^- x{p). 96 ANSWERS AND SOLUTIONS According to equation (9) the first two terms cancel each other. Hence, £{...}—*</». and thus x(p) J V2p(E — U)dx — px(p)=—fx(p)dp. The function G is thus of the form o ( P ) == -^ l e -i /•«• dt and is therefore the same, apart from a multiplying constant, as the function G(p) of equation (7): G (p) = const G (/?), as had to be proved. 2. TUNNEL EFFECT 1. In the region of the metal (x < 0) the general form of the wave function corresponding to the eigenvalue £ is the following: In the region x > the eigenfunction has the form of a travelling wave emerging from the metal <!>„ = ae ikx , where k = T / ^ |x£ -. On the metal boundary the wave functions ^ and <J*ii an d their deri- vatives must be continuous: The ratio of the current density of the reflected wave to the current density of the incoming wave gives us the reflection coefficient If the electron energy is zero the reflection coefficient Ro is equal to unity, while with increasing energy X decreases rapidly; for E :> V vl /? ~ ° TUNNEL EFFECT 97 In the other limiting case E<^v For normal metals V ~ 10 eV. The reflection coefficient for 0.1 eV electrons is then /?o=0.67. 2. In the Schrodinger equation -Irf-fa+nrMl-o (i) 2j* \ e a -f 1 we make the substitution 5 = _ e ~~Z, <|> = r'*°« ($), where £ = X^L , For the function u (&) we get the hypergeometric equation S(l— %)u!'+(\-2lka)(\-\)u'— x*a«« = (y^^). The solution of equation (1) which for x -> oo (£ -+ 0) is finite and behaves asymptotically as a travelling wave ce ika > has the form «|> = <:««**/* {/(x — k)a, — i{* + k)a, \ —2lka, — e~^\ To find the reflection coefficient it is necessary to determine the form of the wave function inside the metal (x-+ — oo): d,« c T{l-2t»a)T(-2tka) ,.„ Y r(— i( % + k)a)r(l— i(k + %)a) e ^ . T(l-2ika)T(2lka) ~ V(l(% — k)a)T(l-\-i(% — k)a) Hence we find for the reflection coefficient d _ I r (2ika) T(—l(% + k)a)T(l— 1(% + k) a) |» _ sh^a (x — k) ° |r(— 2fAa)r(/(* — k)a)T(l + l(% — k) a)\ ~ stf na (% + k) ' To evaluate R a one must use the following relations Y(z+l) = zT(z), T(z)T(l—z) = -^—, K ' v ' sin it,? T*(/A:) = r(— ix). 98 ANSWERS AND SOLUTIONS (x real). As a-»0 the equation for the reflection coefficient goes over into the expression for R in the case of a rectangular potential wall (see preceding problem). It is easily verified that R a < R , that is, that the reflection coefficient in the case of a smooth change of the potential is less than in the case of a sudden change. For a = 1A, V = 10 eV, E = 0.1 eV we find R a = 0.23. 3. We consider a flux of particles with energy E < V moving from left to right. In the region III the wave function is an outgoing wave * BI = C«*». * = -^-. In region I we have both an incoming and a reflected wave fy = e ikx -\- Ae~ iliX . In region II the general solution of the SchrOdinger equation has the form The coefficients A, B v B 2 , C are determined by the condition of continuity of the wave function and its first derivative. At x = this condition leads to the relations: \+A=B x +B t ik(\—A) = x(B 1 — B 2 ). Correspondingly we have at x = a B Y e xa -f B 2 e~ xa = Ce ika , x (B x e %a — B 2 e~ xa ) = ikCe ika . From these equations we find A - C ^ e**« (*" - e-> = C*g2i ,-xa : aika 4ik% " v ' ti* 4/A* B 2 = C- -5.(1 —£)«**«+««, "~~('-?j , -«--('+sr TUNNEL EFFECT 99 Since we have taken for the expression for the incoming wave in fy the form e ikx , the transmission coefficient is D — CC*. Evaluating this we have 4&W D (#> + *2)2 sha %a + 4*3x2 • We note that the transmission coefficient tends to zero when we go over to classical mechanics, that is, as fi-*0. If xa^> 1 fthat is (V — E )^ > -^-^)> tne expression for the trans - mission coefficient becomes extremely simple, h We consider two concrete cases: a) An electron with E = 1 eV passes through a potential barrier with V = 2 eV and a = 1 A. We find for D the value 0.777. b) Now let a proton with the same energy fall on the same po- tential barrier. In this case one can show that the transmission coefficient becomes vanishingly small, D = 3.6 • 10~ 19 . 4. d (& — tf)sln**a (. _ Y2pE _ V2fji(£— V ) \ * 4kW-\- (/fe2_ % 2)si n 2y,a \ h ' x— h / 5. For the case E < V the wave function can be obtained from the expressions for the wave functions of problem 11, section 1, by changing the sign of E and V e . The general form of the wave function corresponding to an energy f is +*.(.bip.hii>(-*+$- 1 4, -X + f + 1,4; -.h.f). (.) where -«(^ ,_&£?_,). "^If 100 ANSWERS AND SOLUTIONS The coefficients c x and c 2 are determined from the condition that as x -*■ + oo the wave function has the asymptotic form To find the asymptotic form of expression (1) we use the relation f(«,^r.*)= r$r(^j <-*>~' f ("■"+'— f- "+'-?= t)+ + f^T^(-«)-'''(M+l-T. P+l—i). We find thus: i>*->-co~(-if{(<vii— ^ 2 )(— \y ka e**+ + (c 1 B l -c,fl l ) (-4)'*"'"**}' (2) + (cA-fc,B 1 )(^) ftfl « to }. (3) where for the sake of simplicity we have introduced the notation r(|)r (-ika) At — r (I) r (- ika) A = Bx '(|)r(l*a) r(_, + -) r (, + i + -)' r(|-)r(£*a) ZJ, = The difference in sign of the coefficient c 2 in equations (2) and (3) is explained by the fact that sh -^ is an odd function and the second term of expression (1) changes sign when we change from positive to negative values of x. The requirement that at + oo there is only an outgoing wave leads to the following connection between c x and c 2 : c x A x -\- c 2 A 2 = 0. The transmission coefficient will then be of the form: \c t A 1 — c 2 A 2 \* ' TUNNEL EFFECT 101 Substituting in this expression the values of the coefficients A v A 2 , &i and B 2 and performing a simple transformation we find finally: D = sh2nka f 0r 8^^ a < j and D = sh»«te 8^0* j 6. The potential energy of the electron is of the form depicted in fig. 10. The transmission coefficient is: where the points x = 0, * = * = I|i limit the region which the r particle according to classical mechanics cannot reach. Evaluating the integral in the exponent, we find: D^e~vT ]E]U (!) To ascertain the limits of the applicability of this result, we note that semi-classical considerations cannot be used near the classical turning points * over a region x — x < (~-\'\ Equation (1) is applied when that region is less than the width of the barrier, x __ \E] x -pr > (^)Vlf or J£|* »1- This requirement is thus equivalent to the requirement that the transmission coefficient is small, D<1. The transmission coefficient D decreases steeply with increa- sing \E\ and increases with increasing F (see table 1). 7. The total potential energy is equal to: <?2 V = — Fx— f-. Ax One must note that this expression is incorrect for small values of x (of the order of magnitude of atomic distances). How- ever, to evaluate the transmission coefficient, the exact shape of the potential in that region is inessential. 102 ANSWERS AND SOLUTIONS 1-t o 1 o o © X in o © E- 2 W ■ i— i u I— I \u w O U z o w (—1 s Z x in x CO X O X X tH o 1 O •fl o 00 o rH d u > 4) CO 1 CO 1 •^ o 1 13 O O o © V rH iH f* rH iH e X X o rH X X m t~ iH c~ CO r-i > o H (J « 5-4 a o u» co *- O CO US bO 1 o O tH o rH bO ctf CO rH © tH O rH a o u a> 'a V 1— 1 ctf O H X X CTS X lO 6 •r-i y O © X CO cvi X CVI •^ •*• £ f) 0a e>9 O "Y* ■ »H <u 1 O 1 O O © o a 9 0) rH iH rH <u rH rH eo O Q) X X X <u X X O , <(-> in U3 <o id CO t- rH OS 3 3 co CO •a tH O •i—i ID 43 £ ? X in X CO X CO X X 00 X 00 s > > > (N co m i i i II II II ha c*3 t«a 1 o 1 O 1 o rH rH rH > > > (N CO lO II II II C*3 t*3 tJd 2 S ■£ 4J TUNNEL EFFECT The transmission coefficient is 103 ( x a \ / x \ ' X * J I X, J The turning points x t and x 2 follow from the condition that the classical particle momentum must tend to zero at x^ and x 2 , p = y r 2 v .(\E\-Fx-£) = 0, X l, 2 — __\E\±V E* — e*P. the integral 2P X t X, X, x t ■Fx—4-dx 4x is a complete elliptic integral. Taking as the independent variable ■— - x = S this integral is reduced to a function of one parameter: f, *«_|/5 1^,00. ,«J^. X, 6i Here <p(v) = i/ l — £ — $</$, and the limits of integration £j and £ 2 are the roots of the expression under the square root sign. Intro* , we get D = e- k &(y). ducing the notation k = 4 V^p ' E ' ' 3 ' -r /? We note that the transmission coefficient becomes D = e- k » If we neglect the electrical image force (y = 0; see problem 6). The values of cp (y) are given in table 2. The influence of the electrical image force on the coefficient of transmission through a barrier can be seen from table 1. Table 2 y 0.2 0.3 0.4 0.5 0.6 0.7 0.8 1 0.9 1 1.0* f(y) 1.000 0.951 0.904 0.849 0.781 0.696 0.603 0.494 0.345 1 0.000 104 ANSWERS AND SOLUTIONS 8. The wave function has the following form: tyi = Asin kx < x < c, ^ u = B l e xa >-{-B i e-* a! , a<x <a + £, <!> III= =CsinA:(2a + 6 — x), a + b < x <2a + 6, where fe = V"2i^ / 2fx(V -£) The requirement that both the wave function and its derivative are continuous leads to the following relations: A sin ka = B x e w -\- B 2 e~ xa , Ak cos ka = x (fl^** — B 2 e~* a ), B x e x < a+& > + B 2 g- X («+ 6 ) = C sin £a, x(B 1 e x ( a+& ) — B 2 e~ % < a+6 >) = — Cfc cos ka. Eliminating B x and B 2 from these equations, we find: (~tgka + l)^*» = (£ tgfta — l)c, From the requirement (jtgfea+l)^ _-(*. t g*a-l) (±tg*a—l)«-«* — (£tg*a+l) = we get (itg*a+l)«'* = =t(^tg*fl-l). This equation determines the energy levels. Using the inequality x6»l, the last equation can approximately be written in the form: tgka = — — + 2 — e- xb . The right hand side of this equation is a small quantity. In zero-th approximation (£<C» we get TUNNEL EFFECT 105 which is the value of the energy of a particle in a potential well (see problem 1, section 1). In the next approximation we have: £ = ^-_A — 2 A e -x j6 (n _ l 2 ~ . a a^-^^a H e l" — f» 2 > o, . . .), The first two terms £<*> = £<°>_ ^ do not depend Qn ^ ^ ^ the approximate values of the energy levels of a particle in the potential well depicted in fig. 26 (b -y oo). In this approximation the levels are two-fold degenerate; this corresponds to the possibility of finding the particle either in region y w I or in region III. If we consider the case of finite b, we find that the possibility of transmission through the barrier leads to a splitting of the levels. This splitting is exponentially small. Let us find in the present approximation the coefficient A, *r B v B 2 and C. The lowest level corresponds to the following coefficients : a *° B l = (— \f- l ^ e -*Ab+a) A> C = A, The upper level *° corresponds to the coefficients C = -A, B 2 = (~l) n ' 1 ^-e^A. *0 -1*0 The value of A which follows from the normalisation condition is equal to -— (in evaluating the normalising integral the contribu- tion from region II can be neglected). We find thus for the lower level the following wave functions: 'W = —p= sin kx, 106 ANSWERS AND SOLUTIONS Y a% o y a For the upper level we find similarly Y~a 1 ko ty L = — ^r sin kx, , ,.n-l 1 «0 f g-x (a;-a) g-x„(a+&-a;)} > "^ y r av. ^ = L_ sin k (2a 4- 6 — *)• 1 y a VY.z; \Za+h x Antisymmetric wave functions (upper substate) 13^ Za+b x Symmetric wave functions (lower substate) Fig. 27 In fig. 27 we have drawn the wave functions for n = 1 and n = 4. 9. For the wave function in the region x< — b we have v VT7I -6 -4- J lPldx TUNNEL EFFECT 107 (the solution must tend to zero at infinity). In the region b < x < — a we have . * x ' V7 V7 —a \ S pax tS pdx = -£=<? ' 4 e ~ b e -j- Yp —a —a c .* i . z b + 7T" In the region — a < jc < -f- a we have _iJL T S pax ('*T "T J 1 * r — e -« 2 Vl/>| a; <!» — eg 4 g _6 { — T=re ~ a Y 12/1/^1 I dx dx ■ -a ir i r IP I <tor l2/|^| ^ -g -« = —j- J u> i ** ' -jj- J |p I fc J — sin [ — \ p dx 1 g -° g * -+- +a + a 2c /l f° \ T J l^ l<fa "t/ IPlfe "7^ cos (t_J/^J - e ~ a e x For -|- a < x < + & we find similarly W ty = c sin I -r- I p dx \ e ~° X 108 ANSWERS AND SOLUTIONS . 4- J" \pldz + 2c cos | jr J pdx\e -* X 9 - l T T J ** r /T ~h) pdx +a , cin I _L I n H v \ i Y~p\ 2 + a -j+6 6 H-2cos(^- ( prfxW -° + 7# sin ¥j " rf *l* - + -f-2cos|| -6 + o +6 +6 da; Finally, extending this solution to the region x>-{-b we have: + a \dx + a 6 . 6 . _lL Up (6 . 6 -i J p dx J sin y J p dx j X + a a; ■i- | |i>|<taj ~L Jiplto + a / f* \ ~T J" |j,|<te H — 4=1 — sin2 1 4- /><**)« "° + +a a i f _ , \ * A i * * 4-4cos 2 (-g- J pdx\e ~° dx 6 In order that this solution tends to zero as x -*■ -f oo it is neces- a; -T- I I i> I dx sary that the coefficient e 6 tends to zero, that is, TUNNEL EFFECT 109 +a e -° + -sin^ljpdxj -f- 4 COS 2 j -r J jOrfX J e -° = 0, \p\dx Assuming the transparency of the barrier to be small, we get the following condition to determine the energy levels: f / 1\ 1 "*"/ "' We shall denote by E l % the energy levels of a separate potential well 6 J" V2v-(.E ( ? — V)dx = Tz(n+±). b h The energy levels in a double well, E n = E^-\-bE n are found from the quantisation rule found a moment ago by expanding }/2^(E n — V) in terms of A£ n and breaking off at the term linear in A£ n +a b __L \v\dx ^i\^,l ■ .."**« - or -4- f |p|*r ac » — 2* where a> is the angular frequency of the classical motion in a sepa- b rate well ^ = 2u f*£. a The splitting of the level E n is equal to 2 1 A£ n |. 110 ANSWERS AND SOLUTIONS in " —a 10. x = —e U) 11. We write the wave function in the region of the n-th poten- tial barrier b n < x < a n+1 in the form — ( \p\ dx ±. C \p\dx dx Y\p\ V\p\ + *n+l J_ f |*> l<ta _J_ f |p|«te if we extend this function into the region of the (»4- l)-th potential barrier b n+1 <x< a n+2 , we get -*/» Ida? * VTTi "T J liM dx I bn+i \ Cfe *» COS 1 J prfxU- w+l ra+i da; / ° M +1 ' + D n e 6 » sinf-g- J pdx + dx -C„* "» f] i*'* X Xsiu a n+l K+i \ tJ lJ,, * r / &n+1 \) J | pdxj-|-2ZV *» cos i J* prf* . We introduce the following notation ±j\,\dx-±j\p\dx-...-i [\p\ax~*. TUNNEL EFFECT 111 The earlier expression for ip in the region of the {n -\- 1) -th barrier is then transformed to the form * e 6 »+i {^e- T cosa-f D n ^sina) + a? + 7J7]* n+1 {— C n «-*sino + 2D B <*coso}=* X x , ~ J IP lite • -i- J ,,,,*, where c n+i = -§■ e ~ T cos a 4- D n e^ sin a , Ai+i = — C n e- T sina-|-2D M e T coso. The connection between the coefficients C n+l and D n+l and C n and D n can conveniently be written in matrix form: C n +i\ f-^e-^cosa «-sino\/C„\ /C n \ A.+J I — «" T sino 2^coso/ \D„/ ~ \D„/' Applying relation (1) AT times in succession we get a connection between C N , D N and C , D , C N \ ./ye- T cosa e- sin a \ /C \ N /C Dj^/ y — £- T sina 2e T cosa/ \D Q ) \D Q The wave function of a stationary state must decrease both for x < a t and # > &# so that we must require that C = Dn = 0. It is easily seen that this entails that the matrix element (A N ) 22 is equal to zero. The condition (A K ) 22 =Q determines the energy spectrum of our problem. To evaluate this matrix element we con- sider the matrix 112 ANSWERS AND SOLUTIONS It can easily be shown directly that the matrix 5 satisfies the equa- tion (2) S->* with the initial condition S(0)=1. We write equation (2) in more detail d_(S n S*\/a p\/5 u S 12 dt\S n S 22 J \i 8/\5 21 S 22 or dt dS 2X dt aS n -\-$S 2v ' = a5 12 + pS 2 dS n ~~Jf — ; * ,J 12 ~T l J °22' -rff — "Wl2 ~T" 8o 22 . Since the condition which determines the energy spectrum can be written in the form <">»=(iH..- ' it is sufficient to consider the second pair of equations. If we put 5 12 =fe xt , S 22 = ge lt , we get The value of A. is determined from the equation a — X p T 8 — X which gives two roots \ t and A 2 * Since X i \ 2 = 1 we can write A 1>2 in the form 0, X 2 — X (2e T + -^- e~A cos o + l*=0, where A l,2 — e ' cosu — le z -\~-j-e~Acosa. If X t =/= A 2 , th e solution which satisfies the initial conditions S 12 (0) = 0, 5 22 (0) = 1 , has the form: M XA S 12 = P(e M — O/^-A.) M. Xt ^22- ^=T 2 ' TUNNEL EFFECT 113 The condition which determines the energy spectrum of our problem can now be written as follows: 04*) 22 = (^P) = -V {(\- <*)Xf -(X 2 -a)Xr) = 0. \ ut I t=0 A.j taj We can substitute in this expression the values: and neglect e~ x , in the equation which determines cos u (this is equivalent to assuming that the penetrability is small): cos u rz. e x cos a. Under this assumption the condition determining the energy levels becomes very simple: sin(AT+l)a ^ sin a This equation has the following roots: nn u= =nTT' with u = 0, iu, 2n forbidden. Thus, cos a has M different values, cos u = cos Trpr^e" cos a (n = 1 , 2 N). N-\- 1 N More explicitly, a, (6, -i- J I p | dx ±r\pdx\=e 6 ' C0S J^n (»==!. 2 A0. Since e b > is a small quantity, this last equation can be re- written in the form: >>i -i J I p I dx lj/>rf* = u(m-f4) + * bl C -N+l (3) .- nn °i cos ■ ' (m = 0, 1, 2, ...) (»=1,. 2, .... A/). 114 ANSWERS AND SOLUTIONS This is the condition which determines the energy levels in the field V(x). It is very similar to the quantisation condition for the field of a separate well. By considering equation (3) we can conclude that the energy spectrum in the field V(x) is, roughly speaking, the energy spectrum of a separate well, with all its levels split into N sublevels. Let us determine the value of the shift, A£ % , -j-fy^cEto)— votf*+g£r dx Ac„, = "a _i_ {\P\dx = *(m-fl)-M * 6 ' cos^C/i^l, 2, .... AO; if we now introduce the notation 6, b, 7t P dx [' dx we find from this a> The distance between the upper and lower sublevels is equal to: a, —4- \P \dx 2ha> h i MB « e bl cos xr i i • 12. In the region x < — 6 we have under the circumstances of our problem only a wave which comes in from -oo, or, -6 '-■^""(tJH- If we extend this solution into the region x>b, we get the follow- ing expression for the wave function: ty = -j==. exp | -^ | p dx J X xUexp(-|J|p|dA : + ii|cosl-l.|p^J + TUNNEL EFFECT 115 b 2 f + 2exp \ T \ \p\dx — i-K cos U- dx X h i I) + ~h exp [~^l pd *){T ex p(-4Ji'i X cos ( -r- pdx\ — /sinl-r- pdx\-{- + 2expl-^-J \pidx\cos I j J pdjf \ a I » -o The quasi-stationary level is determined by the condition that there is no wave coming from + °o. If we put the second term in the last expression equal to zero, we get ctg\±^pdxj = ^t{^txpl-^j\p\dxj + 2txp\^j\p\dxj\ . Considering I — -g- 1 |p|tf*j to be a small quantity, we find that +a / 6 \ 1 jpdx = «{n+ty-^t*p i-^j\p\dx J. from which follows the condition to determine the quasi -stationary DIE) levels En and their width I\ +a (« = 0, 1, 2, ...) -jj\p\dx)=T, 2^ ex P where El E Fig. 28 = K M /*(/£- V) 116 ANSWERS AND SOLUTIONS We find the following value for the transmission coefficient D{E) = = 4exp | * J|p|rf*)cos» |I j prf*)+sin* ^ J ^j . For values of E coinciding with one of the quasi -levels D(E n ) = 1 For I bE I < | E° n | we have D(£° + AE): ra T2 + (A£)2' The behaviour of D(E) near a quasi -level is depicted in fig. 28. 13.* We consider a one -dimensional barrier of arbitrary shape (see fig. 29). Let a particle of energy E > U (U is the barrier height) fall onto the barrier from the left. For x -► ■+- oo we have only an outgoing wave, 4, = Ae i **> (/fe 1 *=iy r 2m(E — U )), (1) ~T and for x -*• — oo a superposition of an incoming and a reflected wave ^ ^eih.w ^Be-ik,? U x = =i/2m£j. (2) Fig. 29 To prove the required relation we use the law of conservation of number of particles (equation of continuity): dt + divy=o, (3) where y= ih_ 2m (^•v<v— «v V4»*> (4) is the probability current density. In the stationary problem under consideration we have ^£. = 0, at so that also div j=0. This equation is for our one-dimensional motion of the form ^f = or js3j x *= const. (5) TUNNEL EFFECT 117 The law of conservation of number of particles means thus that the current density j^tjg, is the same for all x. We evaluate J for * = + ooand x = — oo. Using equations (1), (2) and (4) we find easily: y (+oo) = ^dli, y ( _oo)-=^(l— |B|>). If we equate these two quantities,by using equation (5) we have: 4l|,4|« + |B|«-l. (6) The first term of equation (6) is nothing but the ratio of the current density j in the outgoing and incoming waves, that is, by definition the transmission coefficient D. The second term is equal to the ratio of the values of j for the reflected and the incoming wave, that is, it is the reflection coefficient R. This concludes our proof. 14.* The potential energy U{x) is equal to U (\ +— ) for — a<x<0, to (j Q (l — j) for 0<*<a, and to zero for |jc|>a. If we use the notation (E is the particle energy) ,2 ImE 2 2mU /1X we can write the Schrodinger equation in these regions as follows: Si+*2* BBs0 (l*l>«). (2) S-(*S-*S+*S-s-)* B=a0 <-«<*<: o>. ( 3 ) S-( x o-^-^t)^ = «><*<«>■ (4) Making in equations (3) and (4) respectively the following change of variable, Ht)*"(*°-*° + '°f> < 5 > '-(^fte-*!— *£)■ (6) these equations become |£- "4 = 0. (S'J g-*=o. (O 118 ANSWERS AND SOLUTIONS The general solution of equations of the form (3') or (4') is a linear combination of two Airy functions v and a. Assuming that the particle falls onto the barrier from the left (that is, there is only an outgoing wave for x>a ) and taking the coefficient of the incoming wave equal to unity, we have thus the following wave function :. f|» = £*M 4. Ae~ ik <P 4» = Bu(C)-j-Cv(C) 4. = D«0i)-f-E«0i) d> = Fe ik J° (x< — a), (— a<*<0), (0 < x < a), (x > a). (7) Connecting ty(x) and ^j at the points x = — a, x = 0, and x = a, we get the following six equations for the six unknown con- stants A, B, C, D, E, F: e -ikji _|_ Agikcfi __ Bu <— A) + Cv (— A), / yi («-**.« — ^*"«) = £«' (— A) -J- Cv' (— A), Bu (\i) + Ou (jx) = Da (ji) + £t> ((ji), flu' (fi) -f Ct/' (ji) = — Du' (ji) — £©' (ji), Da (— A) + Ev (— A) = Fe**<fl , Da' (— A) 4- Ev' (— A) = — t Yl Fe ik & . We have used here the notation (8) a a oV' (4 — *o) = (*„*)* f 1 — •§") ' (9) (10) Since the velocity outside the barrier is the same to the left and to the right, the required transmission coefficient y which is defined as the ratio of the current densities for x > a and x < — a is clearly equal to | F |*. Solving equation (8) and assuming the Airy functions u and v to be normalised to satisfy u'(t)v(t) — u(t)v'(t)=l, we find X {[«(!*) u' <-X) - a ((x) v' (- X)P + Mf (I*) u (-X) - u (;x) v (-*.)]>} 1 <[f / (^)" , (-^)-" , (^)f , (-X)]2+X[ u '(ix)a(-X)-^( t x)t;(-X)]2>- X (11) TUNNEL EFFECT 119 Equations (11), (9), (10), and (1) together solve our problem. The functions u(t) and v(t) which are real for real values of t and their derivatives u'(t) and v' (t) have been tabulated. We shall now find the limiting expressions for the penetrability of the barrier in the semi -classical case. It is well-known that one may consider the motion to be semi- classical provided the particle wave length X = ■]/2m[E— U(x)) changes little over a distance Ax~ %. In the regions x < — a and x > a the motion is free and thus also semi -classical {% = ^ = const, so that -^ = o). In the regions — a < x < and < x < a we have respectively from equations (1), (5), and (6): and V y.2 ) iYX' dx dX, dx ~~ 2/C v » . __ ( a V /3 1 d\ _ d% dt] __ 1 V*|/ iYf\' d * dr t dx 2i») a/ »" The condition of semi-classicism | Q. I <C! 1 leads thus to the dx requirement |C|^>1 and h|r>l. (12) We consider a range of energy E<£/ (that is, * < x , jjl>0). We denote the barrier width for an energy E by 21 and we have clearly: '- o -Dr-"('-4)- a - l=a i- < 13 » Condition (12) is, of course, violated near the classical turning points x = ±i (where C = 0, ti = 0). One can, however, formulate a condition of "integral" semi-classicism (semi-classicism in the important region) which ensures a "semi -classical" penetrability of the barrier. For this one must obviously demand that the sections of non-semi -classical behaviour Ax (corresponding to the sections AC — 1, At) — 1) are small compared to / and (a — l). 120 ANSWERS AND SOLUTIONS We have (A#)non- = semicl * k?\ naii- v ^'senricl I dx a\ l L From equations (13), (9), and (10) it follows that the condition ( A *) non-semi cl ^ l leads to tne requirement that ^ ^> 1 and the con dition (A*) non _ semicI <C(a — /) to X ^> 1. We note that to satisfy the inequality X ^> 1 it is in any case necessary that k a = — :> 1, which can be seen from the definition (9) of A if we write it in the form X =(k Q a) Va (^j'\ We can convince ourselves in exactly the same way that for E > U (that is, k»> x , (i < 0) the condition for semi-classicism is the inequality I M ^> 1. and if that condition is fulfilled we have automatically X :> 1 . It can be seen from equation (10) that the inequality | H ^> 1 for fc ^,x n requires in any case that the above-mentioned "trivial" condition for semi -classicism k a == £- 3> 1 is satisfied. Thus, both for E > U and for E < U the semi -classical trans- mission coefficient y can be obtained from equation (11) for | ji | ^> 1, X ^> 1, that is, by using asymptotic expansions for u(t) and v(t). These asymptotic expansions have the following form (we take 2 a/ everywhere only the leading terms; t > and t ^> 1 , x =2 — 1' % ): ■o'{—t), 2 «'(0: ;t v *cos( ^ + -j)» ** ,/4 sin(x + jj; ^ — ^ /4 cosf^-f--x] (14) (15) We evaluate the semi -classical barrier penetrability for two cases: I.- E<U [*:> 1, ^>1- II. E>U M»l (and thus X » 1). TUNNEL EFFECT 121 In case I replacing u(ja) , ... by equation (14) and u{ — X) ... by equation (15) and dropping in the sums the exponentially small terms — e~* (note that taking these terms into account has in general no sense since we have neglected even exponentially large terms ~ — e* with respect to terms ~«* ) we find easily -i.//* "-gE,,^^/. (16) t ^ e s = e 3 nu " As should be true in the semi -classical case t <^. 1 (and y->0 as fi.-^O). Exactly the same result as equation (16) can, finally, be obtained also by using the well-known semi -classical formula t -2 J VlF {V W-W dx 7 = * - In case II replacing a(jx) = a( — \p\), ... anda( — X), ... by equation (15) we get in our present approximation the obvious result y « 1. We note in conclusion two rather trivial limiting cases. If for fixed a and E U -> 0, X -+ oo, p ->• — oo, so that according to case II t - * *• If for fixed o and U E -> 0, X -> so that always ji -► (x a)'/» ^ -+ [ see equation (11)] as in the case of a barrier of any other shape. 15.* If we denote the mass and the energy of the particle by m and E (E < t/ Q ) we have for the transmission (penetrability) coeffi- cient in the semi-classical approximation the following well-known expression : D^expj -2 J y ^-[U(x)-E]dx\, (!) where x t and x z axe the coordinates of the classical turning points, which are determined from the equation U (x 1>2 ) = E, <.-<V>~&- < 2 > Substituting the potential U(x) and equation (2) into equation (1), changing our variables in the integral by putting JL = t and using 122 ANSWERS AND SOLUTIONS 1 the fact that j VT^T* dt=j, we find finally n „f ^/~2^ a(U -E) \ (3) The criterion for the applicability of this formula is that the exponent is a large quantity, '/f fl( V £) > 1 - (4) One can easily verify that if for the barrier under consideration condition (4) is satisifed we can at the same time validly put the factor which multiplies the exponential of expression (1) equal to unity. Indeed, this factor is equal to unity in the case where the field U (x) satisfies the condition of semi-classicism over all sections of the barrier, except those in the immediate neighbourhood of the turning points. This condition is of the form I it* I \dxW2m[E — U(x)V\ Y^m[E — U{x)]> where X is the de Broglie wavelength. Inequality (5) can be transformed to , | <C1 ^| £ (M, p., with t — — . x 2 From equation (4) it follows directly that the condition for semi- classicism (5) or (5') is violated only near t = ±l (that is, x = z*zx 2 ), while the width of the region of non-semi -classicism is of the order of magnitude |A*|==|* — x lt \ ,. a%t>V ° ,. ,. . (6) 11 l,ai m^(U — E) l *U u ^ For the required "integral" semi-classicism of the field U(x) we must clearly have: lAjcKIfl. (7) COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 123 Using equations (6) and (4) we can easily check that condition (7) is satisfied, and thus that the factor in front of the exponential in equations (1) and (3) is equal to unity. 3. COMMUTATION RELATIONS; HEISENBERG RELATIONS; SPREADING OF WAVE PACKETS; OPERATORS 1. We consider first of all the case of a discrete set of wave func- tions ty { . The average values of the operators A and B in the state characterised by the function ty (<j> = 2«»^)> are equal to A = ^a*A ik a k , %, k B = 2 a*B ik a k . i.k We consider the non-negative quantity y(X)= S ^(A ik ^-i\B ik )a k ^ {SMft + AB^a,} >0 where A. is a real parameter. Collecting terms of the same power in X and using the fact that A and B are Hermitean iA ik =A* ki ,B ik =Bli) t we find J (X) — __ {alAMAiM -f ilat (A ki B u — 5 fci ,4 4 ,) a, + i, k,l -j~^a k B ki B iiai \ = J2 + XC-|-X^2. Here C is the Hermitean operator C = j(AB — ba). The quadratic form J(k) is non-negative and thus 4/ 2 B^>(c) 2 . If we note that the operators AA — A— A and AB = B — £ satisfy the same commutation relations as A and B, AAAB — ABAA = /C, we get: ________ The proof of this relation for a continuum set can proceed along similar lines. The expression J(K) — . j {(A -j- Afl) <}>)* {(i -J- HB) <{,} dx, 124 ANSWERS AND SOLUTIONS with positive X is non-negative and can be transformed as follows: y(X) = f [(Ayf—tk(Byfi M> +/xA[>} dx = = / W*A 2 ty + 'M>* (^ — & A ) 4> -+- ^Y^} <* T - Indeed, since the operator <<4 is Hermitean, f (/ty*) cp rfx = f (j/Mcp dx. the rest of the proof proceeds as before. 2. MnVMMN.^! (A<7)*(AF)2>'-i- 3. The energy of the oscillator in a stationary state is c ft/ d/* 2 ■ ** 2 l , / w P 2 i kx* 2[x since and we have p 2 ==(p—p) 2 -Hp) 2 ==(£p) 2 +(p) 2 . ^2=(Ax)2-K*0 2 "2? h 2 fta From the Heisenberg relation (A/?) 2 • (Ax) 2 >— it follows that •c> ===^H o • The expression on the right is minimum for so that c min~ 2 V y — ~2~' where o> = i/ — is the oscillator frequency. 4. In the case under consideration we can neglect the screening of the field of the nucleus by the other electrons. The energy of a K -electron is F __ p* Ze* 2* ~T' COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS i£5 fc since p~—, where r is of the dimensions of the region of the localisation, we have E ~2^—~r- (1) This expression is minimum for r = -~- = -^-( a = 0.529 • 10" 8 cm is the radius of the first Bohr orbit). We have thus for the energy E *.]*= — z *- 13.5 eV. If we take into account the relativistic correction due to the change in mass, expression (1) will be of the form We find thus for the energy 2 E>v. c*{(l—a?Z*y/*—l} where * = ~. 5. Let the regions of localisation of the first and second elec- tron be of the dimensions r t and r 2 . Using the Heisenberg relations we have then for the momenta of the electrons fi h so that the kinetic energy is of the order of magnitude The potential energy of the interaction of the electrons with a nucleus of charge Z is given by and the mutual interaction energy of the electrons is equal to r t -f r, ' To find the ener gy of me ground state, we look for the minimum of the total energy, 126 ANSWERS AND SOLUTIONS The minimum is realised for the values tfl l r, = r- = K 2 7 1 ' Z 4 The ground state energy of an ion with two electrons and nuclear charge 2 is thus equal to A comparison with experimental data shows excellent agreement if we take into account the extreme simplicity of the calculation H~ He Li+ Be + + B + + + C+ + + + £ C aic Ry — 1.125 — 1.05 — 6.125 — 5.807 — 15.12 — 14.56 — 28.12 — 27.31 — 45.12 — 44.06 — 66.12 — 64,8 6. It is not possible. 7. The quantity p is the average value of the momentum of the particle. 8. For the proof we use the operator relation which is valid when the Hamiltonian does not explicitly depend on the time and when there is no magnetic field p=^-{Hr — rH). The average value of p in the state ip of the discrete spectrum is given by p-=^jf(Hr-rH)^d^ Since H is Hermitean, we have l = \ J {«T -r^ — ^rH^dx. Since for a stationary state H^ = E^, /fy = Ety*, we find finally p = 0. COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 127 9. The wave function <|»(jc, t) of a free particle depends on ty(x, 0) as follows: where +oo — CO +oo a(p)= * „ f <K*. 0)e h dx = i{Po-p)x = j7- y(x)e h dx. — CO The function a(p) differs appreciably from zero only for those values of p' which satisfy the condition \Po—P\ 1 ^-i i(Po~P)x Since the oscillating factor e *' changes little provided that condition is satisfied, when x varies within the interval — &<*< + 8, <{>(*. can approximately be written in the form + (x ' 0w «5^ J «<p>«p{{-(j*-£*)}*p (2«fi) - a or 2 _». exp{|-(po^-2^A} / X«P{+T['(*-f')-3F']l*- From the last equation it follows that the wave function $(x, t) will be appreciably different from zero only when the oscillating factor {t[ p (* — ^""Ir']} chan g es little when P changes within the inter- val — j- < p < +y . The dimensi are thus of the order of magnitude: val — -g- < p < -hy . The dimensions of the wave packet at time t ■ 8 +^ 128 ANSWERS AND SOLUTIONS 10. To solve this problem, it is necessary to determine the wave function ^ (x, t), which satisfies the Schrodinger equation ih isr<=m (1) and which at / = has the given value * (x, 0). If H does not con- tain the time explicitly, equation (1) has the solutions E Mx>t) = * n (x)e~ i ~* Lt , (2) where ^ n (x) is the time independent eigenf unction of the operator H fty»(*) = £„<!»„(*). We find the coefficients of the expansion of ty(x, 0) in terms of the set of functions ty n (x), If (X, 0) = 2 0n<K (*)• a »= ff n (*) '> (* > °) dX. n J _ s n . The function 2 a^M (*) * h satisfies equation (1) and is at * = 0- n equal to <|» (x, 0). We have thus E ^(x,t)=^ l a^ n (x)e~ i ~^~ t or where , E n t 0*(«.*) = 2*1(0 *„(*)« h • n To solve our problem, it is thus necessary to evaluate the Green function G t (£, x) and to use equation (3). a) In the case of a free particle, the eigenfunctions are (3) . pr di (r) = - e "»" E — HOL and the corresponding Green function = f__^_V /a ^ (r ~ p)a Since the initial function was «|p(r. 0) = — 1— exp f*£—H) COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 129 we have: whence we find: -Mr, = sr X tbt\ W w<(i+mt Kf) 2 - p ^ 252(1 + ^.\v f* 82 / 2 ^ fi J for the probability density we find: ( (r-m 1 r( i+ ^i)] 1 t+^)1 From this expression it is clear that the centre of gravity of the wave packet moves with a velocity ^-. The dimensions of the packet § t which originally was of the order of 8 increase with time according to the formula *-»/"'+&. p.264 but the distribution in r remains Gaussian as before. Let us estimate the time t during which the dimensions of the wave packet change in magnitude by an amount of the order of magnitude of its original dimensions, T ~-f • For f^>x the linear dimensions of the wave packet increase linearly with time V Let us consider some concrete examples. For an electron initially localised within a region 6 ~ 10 " 8 cm, t is of the order of 10" 16 sec. For a "classical" particle of \x= lg, 5 = 10" 5 cm, t = 10 17 sec ~ 3000 million years. b) The wave functions for the one-dimensional motion of a particle in a uniform field K = — Fx is of the form (see problem 13, section 1), K~Art. [J.0 130 ANSWERS AND SOLUTIONS where 1f B (x)=*A JV^ 3 "*'</«, g^^x+Jj^a, Let us evaluate the Green function: +°° . Bit — oo = i4« jf rf£ g ~ t ^"/Jrf W rft; g -H--^) +i (-- M «) j where Let us first of all integrate over E G t & x) = A* // du dv e- l ^ +i £ +faU - iMX " X + °° . E* / Ft \ X j dEe —00 A9 C C a j -i—+i—+ivto-iuxa 2tiP*/ , /=? \ ~A 2 \\dudve 3 3 i-ll U -\-~ — v\. If we use the properties of the 6 -function we can integrate over v ; we can then put the expression in the exponent in a form convenient for the integration over u , +00 —00 +4-r<*+o+-3r(*-9'}. We finally get for the Green function: °.«-*>-(Tfef«p|--k(£y+ As F-^O this expression goes over into the Green function for free one-dimensional motion. Using expression (3) we can deter- mine the change with time of a wave function given at t =- 0, COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 131 1 ** ,±P0 (*82)V< The result of this evaluation is o fa the general case of a three-dimensional motion in a uniform field with an imtial wave function ^(r, 0) = — L—g-^+T- we get * (r ' ° = f / »mxh/ exp J— A £__Jei/ 1 '« \ , + f(Po+^)r-i-/l£^)i„. o From this expression it follows that the probability density distri- bution remains Gaussian and that the centre of mass of the wave packet moves according to the law of classical mechanics with a uniform acceleration. The change with time in the size of the wave packet is the same as in the absence of a field (see above), c) The eigenfunctions of the Schrodinger equation are of the following form: ^n(x) = c n e 2 H n (ax),. where fte e r r e!at"o„ ed """ ^^ ^ ° SatiSfieS aCCOr<ttn 8 to e< J uati °" < 2 > ^'•^S^wr'T', (4) 132 ANSWERS AND SOLUTIONS where «» = V /W„(«)exp{-a=<i=^ + ifl_i£l} dx. In order to evaluate the a n we use an expression for the generating function for the Chebyshev-Hermite polynomials, e-^^^^^-HM)- (5) X" One sees easily that — is the coefficient of ~ in the expansion Cflfi in powers of \ of the expression +00 —00 From this it follows that: After substituting this expression into equation (4) it turns out that one can perform the summation over n, using again equation (5). If we use the notation we get: a ty(x, f) = cexp{— \\x— Qcos(o)/ + S)] 2 — /*Qa 2 sin(W + 8) — tut ^-l [sin 2 (arf+8) — sin 28]} 2 ' 4 In this case the wave packet does not spread during the motion. The centre of gravity moves as before according to the laws of classical mechanics, performing a harmonic oscillation with amplitude Q and frequency o>. From the expression we have obtained for ip it follows also that the average value of the momentum at time t is equal to P(t) = hQa 2 sin (arf-f- 8), that is, to the classical momentum of the particle in the oscillator. The expression eX p ( — 1?L + ^- 1 [sin 2 (urf + 8) — sin 28]} can be written as follows: COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 133 «'(-f/-^* 11. Let us consider the operator a (s) = e 8 ^ae- 8 ^, where 5 is an auxiliary parameter, and let us find the differential equation which is satisfied by a (s): da(s) j > -> * » * » «. -J—- = Le* L ae-*i< — e^ae-^L s= [La (*)J. We shall differentiate this equation once more, (Pais ) ds 2 P-[L.4g]-aiiHs) U . One sees easily that the derivative *^fi- is equal to n successive commutators of the operator L with the operator a (s). If we now write the operator e^ae-^— a{\) in a Taylor series, and express the derivatives with respect to s at s = in terms of successive commutators of the operator L with a(0) = a, we get the relation which had to be proved. 12. The Hamiltonian operator can be expressed in terms of the operators a and d+ (see problem 5, section 1), H= fro (a+«+ 1) _/(f)j/"JL (a + i+) and we shall write the solution of the SchrSdinger equation in the form $ (t) = c (*)■«« <*> « V (*> « gT (*) o +a y (_ oq), To differentiate the operator acting on ty ( — 00) with respect to t it is necessary to bear in mind that a and a+ do not commute, aa + — a + a = [a, a + \= I. 134 ANSWERS AND SOLUTIONS The expression for the time derivative is transformed in such a way that it has the form where G is some operator. Consider first of all the third term on the right hand side of the equation for ip . Using the fact that [(a+) n a]= — n(a+) n - 1 , we find = (a — a)(l + aa + +^^-+ .. .) = (a — a)e«« + . The third term can thus be written in the form 4 (a- a) ^ V V" + >(— oo). Similarly we can transform the last term, = e-f ( a + a — aa + -+■ & — a?) e a<i V" e T " + ° . We thus have G = c-j-a+a + c (a — ocj) a + -f c({5 -f p-f) a -f- c — cfa^ and in order that the Schrodinger equation be satisfied we must require that ihG = rt. Comparing the coefficients of the operators a + a. a, a + and the con- stant terms, we get the following set of equations: Y — — iU) > *+ iws= vm m Solving the equations with the initial values a( — oo) == 0, p ( — oo) = 0, | c (— oo) | = 1 , we find: COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS t . .. Ip—Usst f 135 — oo t m-7§=J/<^-~*--*<o. T(0== — iwt, c(t) = e 2 ex p|~2/^ \dt'f(t')e-M jV/(f')* tW " The probability for a transition from the state <},(— oo) to the n-th excited state at f = -|-oo is equal to Ur B =lim |to(0^| a ==lira |cfae^V°V 5+<i <K-oo)rfjc|\ If the initial state is the ground state <}>(— oo) = ^ , we have, since Moreover, since the normalised wave functions are of the form (see problem 5c, section 1) % _<« + )' +0 > we have In view of the orthogonality relation f£<!( m dx = Z nm we get: I « (0 |*» W^^limlcCOl 2 - *->oo n\ From the equations for a.[t) and c(t) we get: lim|a(0| 2 = -H^- ->oo ' 2fi{X<0 *->oo J/0 e iuit dt (+oo 2\ -27^ J/(0**-* J We thus find for W m a Poisson distribution with W n0 = e-^ r 2fifxoj +00 ,iu>* ^ a) For /(*)=/<>« * we find w 2,.2 <oM 2fl(xu> 136 ANSWERS AND SOLUTIONS b) For /(*)=-— Jj* — we find (4) + 1 «Yo .2/2-2 -2u>x 13. The SchrSdinger equation for an oscillator with a perturbing field is of the form d<\> _ #» tfty . fxo>2.*:2 < s w = 2??^ + J1 2-*-/«^. Introducing a new coordinate x t = x — $ (0» we have "f-^-^^+yi-'C.+O'WWC+Bt. If we put ^ = exp(^)cp(^/), we get for <p the equation where L is the Lagrangian 1 = ^^ — -jV^SH-fiOb hi the last expression the term (pi-|-[Mo 2 $ — f)x t (f is equal to zero because £as a function of time satisfies the classical equation of motion of an oscillator acted upon by a perturbing force, If we introduce still another function x by cp = ^exp I -^ Id* J, we get for the function x an equation which is the same as the equation of motion of a free oscillator <?X ft 2 (Px (J.a) 2 ATj X- dt 2(* dx\ The wave functions of the oscillator acted upon by a perturbing force can thus be put in the form W*. t) = x\x— 5(0, *]exp J^.i(x—^-\-±JLdt\. o j 14. We shall write the SchrSdinger equation dt 2jx dx* ' 2 T in the form <},(*, /)== f 0(x, t\ x f , i)ty(x', i)dx'. COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 137 One can easily show that the Green function 0(x, t\ xf, x) must satis- fy equation (1) and the initial condition Urn 0(x, t; x', x) = 8(jc — x / ). (2) We shall try to satisfy these two conditions by putting Q(x, t; *', x)~exp[^[aif)x*+2b(f)x + c(fi]}. (3) Substituting expression (3) into equation (1) we get the following equations to determine a, b, and c 1 da a 2 ,,,* |x dt j* 2 w db a , dt H> p. " (4) The solution of the set of equations (4) is of the form t fl = jxJ., £=.£21151, c = ih\nZ — -jb i dt, (5) where Z is a solution of the equation We try to satisfy the initial conditions (2) by suitably choosing the integration constants. To do this we use one of the possible expres- sions for the 6 -function, >c»-*o- 'J;;/" *,„,;;_,) «p(g^c-*y } <6> (see problem 10a, section 3). In order that expression (3) for t -* x goes over into expression (6) it is necessary and sufficient that Z = 0, Z = 1 for t = x, o— z , c = ih\nZ-\-px' -y-, where Y is a solution of the equation y = — u> 2 (t)Y with the boundary conditions Y=l, Y — Q for * = x. We note that since ZY — YZ = 1, we have JL — — [ *L Z~~ J Z»* 138 ANSWERS AND SOLUTIONS We thus find for the Green function of our problem the following expression: 0<*. t; x', ^/^exp{^(Z**-2^+r*' 2 )}. For the case where w = constant, we have: Z=-sinu>(/ — x), F = cosu)ft — x), and the Green function in that case is equal to: G(x, t; x', t)== = V 2nhl sZ (t - x) ^ { 2h JT(t - x) < C0S "Q-x)*- — 2xx' -f- cos o) (t — x) x' 2 )} . 15. j/ cos8arf + — sin»arf Xexp where c&jx— Qcos(^ + 5 )) 3 ) a*h* I ' «*'"*+ ■p£* sln ' orf J " ' [AGO ^ In the case where a = y -^ , we get the result of problem 10c, section 3. 16. Q (x, t; x', t) = -/^«p{A |i(x -° l - 2(JC - S) *' +|r f ,+ In this expression 4 satisfies the equation y& = — jx<o 2 ^ -f-/(0 and the initial condition &(t) = 0, £(t) = 0, and L is the Lagrangian 17. The probability of a transition from the state n to the state m is given by the relation (we use the system of units where h = l, t*=l, (0 = 1) P mn (t,0) = \G mn (t, 0)|«, (1) where 0«»('. 0) = fftf m (x)0(x, t; x f , ty^ix^dxdx'. COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 139 Using the generating function exp j —^(2* + * — 4zx) } = 2 ]/" ^p- «"*„(*) n=o we can construct a function G(u, f) G(a, v)= J J expj — y(2t> 2 + ;t 2 — 4vx) — _^.(2 u 2_|_^ 2 _4 ajc /)j G(x, *; *', 0)dxdx' = CO CO = 2 S/J^^O^.O). "srsr ™" v ' <2) t»=0 n=0 * ' From expression (2) it follows that the quantities Q if (f t 0), the absolute squares of which determine the transition probabilities, are apart from a factor y n2%+ . the coefficients in the series expansion of Q(u, v) in powers of u and v. Let us evaluate Q(u, v); to do this we substitute into equation (2) the expression for the Green function G (see problem 16, section 3). After this substitution we get + ( l -^ x >*- 2 (2v-§l+il)x-2(2u + §-)x']}dxdx'. When w = constant, S, Z, F are of the form t Z(t) = f sin (t — t')f(t')dt' , Z = sint, K=cos/. o To evaluate the integral we use the formula + CO J I dx dy exp { — -i (ax 2 -\- 2bxy -f- cy 2 — 2px — 2qy) ) = — CO = 2n exDi ag2 ~ 2bpq + Cpi ) Yac—tfl \ 2{ac — b*) ]' After some simple calculations we find the following expression for the function G(a, v): G (u, v) = V* e iF(t) e~~* exp { — ^£_ uv -f Au -+- Bv } . 140 ANSWERS AND SOLUTIONS where t A = i fe-K'fWdf, B^e-^A, 2w = \A\-\B\=\ 2 -\-?, o and F(t) is some real function of the time. To expand G(u, v) in a power series, we use the relation «p{.+p-4j- 2«c«i->-S5. , m, w=0 where min (m, n) „/ i \ V' mini , ._7 c(m,n\w)= 2, l H m-l) H n-l)S - w ) ' 1 = Expanding, we now get G(u,v) = V-e iF %~ V c{ m> n\w) {Bv r {Au r- +•+ ' ' ml n\ m, n=0 From equations (2) and (3) it follows that (3) G mn {t, 0)= ' 2 ' AnB =c(m, n| «)«<*«, V2 n+m -m\n\ and the required transition probability is equal to For the particular case n = the transition probability is given by P m0 (t, 0)= e ~ w ^» m t S i nce c(Wf o|w)=l. After we have evaluated the transition probability we can determine the average value of the energy and the square of the energy of an oscillator at time /. These average values are given by the equations oo oo m=0 To evaluate the two sums which are similar in character we consider the expression [l — — ) e a = $(m, u\w). COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 141 One shows easily that CO $(m, a\w)= V — rc(m, n\w). _ n=0 From the equation ~ $(m, a|w)$(m, p|w) = e a P /M ' 2- ml t» = it follows that CO 1 e -w. w m ml C ( m »» = c (w, n\w)- c (m, n'\w) = &„„-«! w~ n , so that the physically obvious relation, oo -Zj 'mn == * w = follows directly. Consider now the equation 00 V m e~ w -J m $( OT>ct | w ) $( m> p| w ) = { w _ a _p_|_^j e «p/w > l» = and differentiate its left hand and right hand side n times with respect to a and m times with respect to and after that put a = 13 = 0. We thus get CO 2 mP mn = * + •*>• «»-0 In this way we see that the average value of the energy of the osci- llator at time t is equal to E = E n -\- w. In this expression w is the work done by the force f(t) over a period t , t t w=J/(/)id/= \(\-\-l)\dt = In a similar manner we find £2 = 2w£ n . 20. a) i(f)- x -!L t £.. b) jc (0 = JC cos o)^ — — -sino)^. 21. (A*)* = (A*)* + 1 [(Ap) (A*)-h (Ax) (Ap)] + -J (A/>)*. 142 ANSWERS AND SOLUTIONS Note . From the relation we have derived one can easily determine the moment t when the quantity (A*) 2 has a minimum value. The function (Ax) 2 , is symmetrical with respect to the point t. In the case where the wave function at t = has the form <J> (x) = cp (x) e h (cp (x) is a real function) t = (see problem 10a, section 3). 22. a,(^ = (A^. + ^f (£JV — OO +oo b)(A^ = (A^ =o . cos 2^_ f _^ J (|j) 2 sin2a>/^. 23. * (AB) C — C (AB) == ABC — C,4B -f- ,4C5 — ACB =~A(BC — CB) -\-(AC — CA)B or in more compact form [AB, C\ = A [B, C\ -4- [A, C\ B. (1) Since [L v L 2 ]== — [L 2 , L t ] the identity (1) can be transformed to the form (by keeping the order of the letters A, B and C the same while renaming the operators) [A, BC] = [A, B]C-{-B[A, C]. (2) We emphasise the similarity of the equations obtained, especi- ally equation (2), to the formula for the differentiation of a product of two functions, which is also identical in the order in which the operators which form the original product [A and B in (1), /} and C in (2)] occur in the different terms. 24.* [24 24]-(24)(24M24)(24)=2244- * k * k k i i k 2 2 B k A { = 2 [A t , £*] as had to be proved. i k i, k 25.* We shall denote the operators under consideration by A and B and the Hamiltonian of the system by ft. We have then: After an obvious transformation, using the results of problem 23, dA dt dA • section 3, combining the corresponding terms in -7r— A and COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS I43 ?afiwe find the required rule for differentiation at 26.* In the formula the denominator is, of course, positive so that we must prove that the enumerator is positive. Since L is Hermitean, we get: J* tfLLAidx = J (£<}>) (L*<f) dx = J | b) | a dx > 0, as had to be proved. 27.* 1) It is not linear. Indeed, denoting the operator of com- plex conjugation by K, we have K (ci'W + ^2) = tf &W + clK^ * c,k\ + c 2 ^ 2 . 2) It is not Hermitean. Any Hermitean operator satisfies the condition J* <5»*I<P dx = j x?L*<fdx. (1) In the present case we have so that condition (1) is not satisfied. 3) The operator K is its own complex conjugate, that is, it is real. Indeed, so that Denoting ^* by f we have: /C*cp = cp* = /Ccp, or fC = K. (The operator of complex conjugation changes / to — /, and its complex conjugate changes — / to /, which is the same). 28.* Let us write the required expansion as a power series, CXI n=0 144 ANSWERS AND SOLUTIONS where L n is an operator to be determined. We have thus ^ n-0 Multiplying by A — IB, for instance, from the left we get 1 = 2 X n (A — kB) L n = At Q + S k n {AL n —BL n ^). n-0 «=l Comparing the coefficients of the same power of A on the left and on the right we get: AL =l, AL n — BL n _ x = 0, whence ^ A aa Lq = /l , L n = /l tiL n _ l . We have thus In the case where » and B axe numbers, this expression goes over into the usual one 29.* It is, generally speaking, impossible to give a definite answer to the question on the basis of commutability or non- commutability of the operators A and B only. Indeed, on the one hand, commutability is, strictly speaking, not necessary in order that the corresponding quantities have simul- taneously well-defined values. For the motion in a spherically symmetric field, for instance, we have simultaneously well-defined values (equal to zero) of all three components of the angular momen- tum for all states where the square of the total angular momentum is equal to zero, although any pair of these three operators is not commuting with one another. The non -commutability of the operators A and B (first case) does not exclude the simultaneous realisation of some of their eigen- values. On the other hand, from the fact alone that A and B commute (second case) it is, generally speaking, impossible to conclude that the quantity B has a well-defined value in the state W A . The fact is that commutability of A and B is sufficient only for the existence of a complete set of states with simultaneously COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 145 well-defined values of A and B (we shall denote the wave functions of such states by W AB ), but not to guarantee that any given wave function W A belongs to the set of functions W AB . This is connected with the phenomenon of degeneracy, where one value of A corres- ponds to a number of values of B so that the wave function W A is in the general case a linear superposition of functions W AB of the form which reduces to W ^ only for a very definite choice of the coeffi- cients (c B , = o B , B ). We shall consider two simple cases: 1) For the motion of a system of particles in a spherically symmetric field in a state with a well-defined value of the square of the total angular momentum M* the value of its z-component M z can either have a well-defined value, or not. Indeed, all non-zero values of M* axe degenerate with respect to different values of Af z (this is connected with the existence of the operators M x and M y which commute with M 2 , but not with Mg); Wm> is thus a superposition of the form fin = 2 cm ^m*. m ■ M z g z In particular, one can find such states W M * where M z has definitely not a well-defined value, notwithstanding the fact that the operator M g commutes with M 2 ; such states are those with well- defined values of M 9 or M y (except for the state where M a = 0, when also M x = M y = M z — 0). 2) Each energy level E , except the ground state level, of the hydrogen atom corresponds to several values of the square of the angular momentum M 3 so that the states with a well-defined energy are, generally speaking, not characterised by a well-defined value of Af a , but are a superposition of states with different values of M 2 > and only for a definite choice of the coefficients of this super- position do they reduce to the functions W B , & . From the fact that E is well-defined it does not therefore follow that vW a is well- defined in spite of the commutability of M* with the energy operator H. The commutability of A and B therefore does not in general yield an answer to our original question. To solve it we must let 146 ANSWERS AND SOLUTIONS the operator B act upon the function W A . If the equation BW a = Wa, is satisfied ( \ is a number), W A is a wave function of B and the quantity B has in the state W A a well-defined value (X); in the opposite case B has in this state no well-defined value. Only in the particular case of no degeneracy (for instance, for a one -dimensional motion bounded at least at one side) it follows automatically from the commutability of A and B that any eigen- function of A is at the same time an eigenfunction of B , and vice versa, so that for any state either both the quantities A and B or neither of them have well-defined values. 30.* Let us denote the totality of the coordinates (or, more generally, the totality of the independent variables of the chosen representation) of the system by one letter x, the Hamiltonian of the system by H(x) and its wave function by 4/(x). Let the trans- formation under consideration change the set of coordinates x to the set x'\ x-+S. (1) Let 6 be the operator of the transformation (1). The invariance of the Hamiltonian with respect to this transformation means that H(x') = H(x). (2) Let 6 act upon H). From the definition of 6 and equation (2) it follows that: OH (x) <b (x) = H (*') 6 (.*') = H (x) -> (*') = H (x) 6^ (x), which is equivalent to the operator equation 6H = HO. (3) We introduce the operator O" 1 which is the inverse of 0, where 1 is the unit operator, that is, the operator of the identical transformation. Multiplying equation (3) from the right by 6~ l we 0H0 - X = H. (3') This relation, which is identical with the commutation relation (3) expresses the invariance of H under the transformation 6 in operator form. COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 147 31.* We are concerned here with finding the combined integrals of motion (excluding parity) of a system of interacting point particles in given external fields. In classical mechanics this problem is solved by finding the coordinate (or time) transformations under which the Lagrangian of the system under consideration is invariant. In quantum mechanics the solution of this problem is formally even somewhat simpler: the integrals of motion are those mechani- cal quantities, the operators of which commute with the Hamiltonian of the system (and also which do not depend explicitly on the time). The commutability of those operators among themselves depends in general not on the form of the external field. The pairs M x and M y , M x and p y , p x and / (/ is the inversion operator) and so on, do not commute, while M x and /, M x and p x , and so on, do always commute. We shall restrict ourselves to the case where the external field can be derived from a potential (excluding thus magnetic fields, and so on) and we shall number the particles by an index /= 1, 2 N. We can then write the Hamiltonian of the system in the form (/, k=l, 2 N) i=l + ^ext (n rtf)=s// -|-£/ ex t, where U int is the potential energy of the interaction between the particles in the system. All operators of interest to us, namely: P(P X , P y , P.), M(M X , M y , M s ), M\ I, commute clearly with H (which corresponds to a bound system of point particles; the non -commutability of the operators lti x , lti yi among themselves leads clearly to the degeneracy of the energy eigenvalues and the total angular momentum eigenvalues with respect to these quantities, Ma, , ) and we thus have to evaluate only the commutability with the operator of the potential energy of the external field, N U ext =St/iext W- * = 1 We shall write down A = const for every integral of motion A Du.,,-0. *» = o. 148 ANSWERS AND SOLUTIONS The integrals of motion of the total system are: the energy E, the three components of the angular momentum M x , M y , and M g , the total angular momentum Af a , the three components of the linear momentum P x , P y , and P x (that is, P) and the parity /. The energy levels are degenerate so that each state of the system is characterised by a complete set of a smaller number of (commuting) mechanical quantities, for instance, by or E, M a , M s and so on. A similar situation also holds when external fields are present. 2) Take the z-axis along the axis of the cylinder. so that the operators p = ly * M,= ^yJ- and/ commute with t/ ext (and thus with H). Since -^ = 0, we get finally E — const, Pg = const, M z = const, /= const. 3) Take the x^-plane in the plane. The operators N u ext = 2^(1^1). •£-<>• N ^ i JU dx 4 ' t=i 7? # __ fi Y d _ 2L Y f J? JL^ and also / commute with U ext (that is, also with H). We have thus £ = const, M„= const, Pa, = const, P y = const, /= const. 4) Take the origin at the centre of the sphere. COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 149 The operators M x , M y , M e , M* when expressed in polar coordinates act only upon the variables fy, <p* and commute thus with U ext and H. Hence: E~ const, M x = const, M y = const, M g = const, /W 9 = const, /= const. 5) Let the half -plane be the x v-plane on one side of the v-axis. Thus we have E = const, P y = const. 6) Let the points be along the z -axis. We have thus: £ = const, M, = const. In the particular case where both points carry the same charge (both in sign and in magnitude), u i = £/ i (p i , |z 4 |), and parity / is thus also conserved. 7) Let the field be along the z-axis. Denote the field intensity by g(0 and the charges of the par- ticles in the system by e it so that we have N t/ext (*) = — *(0 2 Wl (the same expression holds for the gravitational field, if we replace 8 by £ and the charges e* by the masses m^. N U ext commutes with M g ==j^ Xi -~—y~y i-l N N p - * V d p _2V_L Hence M t = const, P x = const, P y = const. 8) Take the ^-axis along the axis of the conductor. w- w-/(oiu <w . ja.-4S-4r- ^.-tS^- i=l i-l i=l In this way we get M s — const, P. = const, and also / = const. 150 ANSWERS AND SOLUTIONS 9) Take the origin in the centre of the ellipsoid, so that we get u ext -S^d^i, 1*1. i* 4 |), since the ellipsoid possesses three, mutually perpendicular, planes of symmetry. The Hamiltonian H commutes thus with the reflection operator / (and does not contain the time explicitly). Hence, E= const, /= const. 10) Take the z -axis along the axis of the screw, let its pitch be a and denote by <p the angle of rotation around the axis. N The function Uext = 1Z UtiPv ?«• **> is invariant under the trans- formation <Pi-+<Pi + 8 ? 0=1.2 A0, ^i->^i-f"2^ fl because 89 = 2u should correspond to Iz = c, with P< fixed, In other words, the operator U ext (and thus the Hamiltonian tf) commutes with the operator i i We have thus M z +-?-P z = const, and also E = const, since _=0. 11) Take the lateral edges of the prism in the z-direction. N N Uext = 2 u t (* 4 , y$. 4r = 0. ^ = 7^' iTi *=* so that E = const, P, = const. In the particular case where the cross-section of the prism is a regular polygon of order 2/t (with n an integer) the prism possesses COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 151 two mutually perpendicular axes of symmetry so that U ext = ==. 2*^(1 ■*<■!• I .yd) (the z-axis is taken through the centre of the polygon) and because / and H commute, parity / is conserved. 12) Take the z -axis along the axis of the cone. If the cone is a single one, N Uext = 2^i(Pi> *i). i-1 so that E = const and M z = const. If the cone is double we have (the origin is taken at the vertex) l/ext «2</«<P«. I*« I). and therefore also / = const. 13) Let the z-axis be along the axis of the torus and the X y- plane in its equatorial plane. We have in cylindrical coordinates N Uext ==2 U^, |**|), and thus E = const, M g = const, / = const. 32.* The error in the reasoning given lies in the identification (as to order of magnitude) of the precision Ax with which the par- ticle is localised in space with the dimensions of the potential well a. In reality the precision of spatial localisation of the particle is determined by the "width" of the coordinate probability distribution |<K*)I 9 ; (similarly the indeterminacy of the momentum is given by the width of the momentum probability distribution \Q(j>)\ 2 ). From a consideration of the normalisation integral f l^pdx it follows that the probability to find the particle inside or outside the well is respectively w and (1— xa), where xa = -^¥^<Cl. Using a rough classical analogy one can say that the particle spends by far more of its time outside the well than inside it. This is a typical quantum effect since the region outside the well is classically inaccessible. We can use the deuteron as a three- dimensional example; its radius is appreciably larger than the range of the neutron -proton interaction, so that the particle spends a considerable part of its time outside the potential well. The effective width of the coordinate distribution is of the order of magnitude (compare problem 4, section 1) 152 ANSWERS AND SOLUTIONS (1) On the other hand one can show that the effective width of the momentum distribution is equal to (compare problem 3, section 1): (A/O ef f~ fi *- (2) Multiplying equations (1) and (2) we get, as should be the case, the Heisenberg relation, 'Ap) eff (Ax) ef{ ~ft. (3) Let us now evaluate more precisely the coefficient of h, in equation (3) by evaluating the product of the average square fluctua- tions in momentum and coordinate: Ap=s=V (p—^ and A* = V{x^xf. ( 4 ) We have, of course, (J>—P)* = P*—W* < 5 '> {x—~xf = x^—^cf. < 5 ") From the known wave function it follows that x = j x[4t(x))*dx = 0, — oo oo p= j p[G(p)]*dp = 0. — oo If we assume a 3 <C! -4r- one can easily show that (6) 00 ^3- *»= J *M4»(*)l a <**«13F- ( 7 ) ~00 The quantity °° J*= J* p^lOip^dp — oo can also be readily evaluated in the coordinate representation, that is, by using the equation oo oo f= ^(x)p'Mx)dx= J <K*) (7 ^j)'* (*)<** = -00 —00 00 COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 153 Integration by parts simplifies this equation to dx '-*l$ rf|\a dx) — CO and by substituting the wave function we find, to the same approxi- mation, ^»av. (8) We note that this value of p 3 does not at all correspond to the kinetic energy in the classically accessible region (that is, inside the well), U -\E\^U , but to the much smaller value | E |. This result could have been expected qualitatively since the particle spends most of its time in an inaccessible region (where its "kinetic energy" is negative), which leads to a large negative contribution to the average kinetic energy (E^-U) = A-p*. From equations (4) and (8) it follows that Ap = fa, Ax = 1 and thus V~2% ' A/7 A* = — . It must be remembered that the least possible value of Ap Ajc is equal to fi/2 corresponding to a "Gaussian packet" <^^e~ b£> 33.* We shall denote the first of the required operators by M&) and have thus: in the r -representation (— = — X —ty( r )=y (r), (1) in the /^-representation (1 = ^ M {p) g(p) = /(/,), (2) where g(p) and f(p) are the Fourier components of the functions ty(r) and <p(r) (the wave functions in the /^-representation) which are connected to them as follows : l f — ^ g(P) = -^%J* <r,C " *• (3) m = -^i' f(r)e ~''^ dr - < 4 > 154 ANSWERS AND SOLUTIONS Apart from here, it is also necessary for other problems to know the Fourier transform of -J- . We shall find it in the simplest possible way. We write: 1= \a(k)e ikr dk. (5) Operate on both sides of this equation with the Laplacian A = V 2 . We have clearly the identity A (<?"»•) = — k*e ikr , (6) and also A(lj = — 4k8(/0, (7) where 8(r) is the <5 -function. Equation (7) is easily proved by writing A as div grad, integrating both sides over a small volume and using Gauss's theorem. Applying the Laplacian operator on to equation (5) leads then to: _ 4u8 (r) = — J &a (k) e ikr dk . From the Fourier theorem and the definition of the 5 -function it then follows that &a (*) = —^ J 4 * 8 (T) «-<*•■ dr = ^ . so that and finally l _ l f e ikr <tk (8) r ~ ' 2w> J #» • To determine the form of M (p) we substitute expression (1) into equation (4), use equation (8), substituting * = 4- in it. We then get: l T l --^ __L_.'ffm. 4 ■—'"*'*. (2nh) I* 2n2 ^ J J W Integrating over r and using equation (3) we have: COMMUTATION & HEISEN3ERG RELATIONS. WAVE PACKETS. OPERATORS 155 Choosing p — p' as a new variable of integration and express- ing fiP) through equation (2) we get finally: (it is clear from the derivation that the integration is over the whole of p' -space). We see thus that - s= M&) is an integral operator with the kernel In a completely similar way we find for the — operator in the r -representation, which we denote by L(r* the equation krMr) =>■%&) (r-r')*' that is, [-]sL (r) is an integral operator with kernel 0(r. r') = 34.* Let ty(x) be a function, acted upon by — (we shall here and henceforth drop the index x of/?). Let <f(x) be the result of this operation, *«7-*- (l) Our problem is now to find a way to determine cp(jc) for given ty(x). To solve this problem we go over to the momentum represen- tation. Let f(p) and g(p) be the functions corresponding to <J». and cp in the momentum representation, so that equation (1) can be written in the form g(p) = jf(P)- (!') Equation (l f ) shows that the function g(p) has in general a pole at p = 0, so that it does not satisfy the general requirement asked of the wave functions in quantum mechanics. In order that this require- ment is not violated, it is necessary that /(p) tends to zero for P=*0: /(0) = 0. (2) Since f(p) is the p-th Fourier component of the function ^(x), we can write this condition in the form J <[>(*) dx = 0. (2») 156 ANSWERS AND SOLUTIONS We shall assume in the following that condition (2) is satisfied, because only then can the operator — be determined uniquely. To solve the problem it is necessary to transform equation (l 1 ) to the coordinate representation. We multiply thereto equation (1») . px by — and integrate over all values of p: <?{x) = -L= f fiPL e -ir d (3) T ywn j p y — oo It is now convenient to consider the integration in equation (3) as an integration along the real axis in the complex p -plane. Since the function under the integral sign does not have a pole at p = by virtue of condition (2), we can displace the contour of integration in the region of this point downwards (to the region Im p < 0): <P (x) = — Lr f ^M e~ r dp. * W Y*rt J. P (3 f ) We now express in this equation f(p) in terms of <]>(*): . f ipx f(p) = -±=r J f (*')«" " dxT. (4) —00 Changing the order of integration, we have 00 pi — <x—x') fW-aJtc^J^T-* < 5 > -00 "v Evaluating first of all the integral over the momentum in equation (5) we have i £-(«-*> J iL - r - dp. (6) The value of this integral will depend on the sign of (*-*'). For x > *' we close the contour by a semicircle of infinite radius in the upper half -plane. The integral over this half -plane is equal to zero, so that the integral (6) is equal to 2it/ times the residue at the only pole p = 0, which lies inside the contour. The residue is equal to unity so that <1L *»-*') J £ — p dp = 2rct (x > *')• (7) or For x < x' we close the contour with an infinite semi -circle in the lower half -plane. As there are no poles within the contour we get clearly ■ p t -=- (X-X') J £_^ dp = (x<x'). (8) COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 157 Substituting expressions (7) and (8) into equation (5) we find X ?(*) = -! ^{x')dx' t (9) —00 which is the solution of our problem. Similarly — is an integral operator of the form T = T /<*•■ < 10 > , -—CO This result is natural since we looked for the inverse of the mom- entum operator p = T ^-. indeed, letting the operator p act upon equation (1), we have fa (*) = P — <|» (*) = 4» (x). (l l) Verifying this result by using equation (9) we find r x ~\ *- — oo -J It is necessary to note the following regarding equations (9) and (10). The contour of integration in equation (3) can be changed also in a different manner, for instance, displacing it into the upper half- plane. In the corresponding evaluation we are then instead of equa- tion (9) led to the equation oo <?W = -iJH*')dx'. (9f) Expression (9') for <?(x) differs from expression (9) by j- jW)<**'. (12) which vanishes for functions satisfying condition (2). The definition (9) of the action of the operator — is thus the same as (9») and is generally unique. When condition (2) is not satisfied, the definitions (9) and (9 f ) are different which shows the non-uniqueness of the action of the operator — . Similarly we find easily for the operator -1 in the ^represen- tation the following expression T~— ff \ dp '' (13) — oo This operator is thus also an integral operator; the operator equa- tion (13) is equivalent to the relation ~f{p) = -{ jf(p / )dp f . — oo 158 ANSWERS AND SOLUTIONS 35.* We choose our origin at the left hand edge of the well; the wave functions of the stationary states of the particle inside the well are then of the form where a is the width of the well; outside the well ^ = 0. The required matrix elements of the coordinate x are equal to oo O X mn — J <& (*) ^n (*) dx *" J *M» djf • (2) -So o Substituting expression (1) into equation (2) we find after a simple calculation 2 1 T T n(m-n )x *{m + n)x ~\. . =s -5--tJ x L c08 — s « J (3) llcosr«c(m + n)l l-cosf^m-n)] !^ K~ 1> OT ~ W -*I OT " From equation (3) it is clear that the only non -diagonal (m±n) matrix elements of the coordinate which are different from zero are those with an odd difference («-n), that is, the elements corresponding to transitions with a change of parity. (We remind ourselves that all wave functions (1) have a well-defined parity, namely » - 1, 3, 5, . . . have even and n= 2, 4, 6, . . . odd parity. ) The diagonal ele- ments (m = n) are different from zero. One can find their value, for instance, by letting(» — n) -+ in expression (3), and using the fact that 1— cos *■»-£-** . The result is a (This result is immediately obvious, since x mm is the average value of * in the m-th stationary state which is equal to %a). The selection rule obtained can be visualised by the following simple considerations. The matrix of the coordinate x is the sum of the matrices of the quantities x -\a and \a . The elements of the first of these matrices are different from zero only if the product M„ isofoddparity , thatis, forodd (m-n) since *-£ a is an odd function with respect to the centre of the well. The elements of the other matrix are clearly equal to -f-8 mn , that is, it is a diagonal matrix. The selection rules found follow thus immediately. To find the matrix of the momentum (p x ) mn ^Pm« we use the well-known relation, Pmn = H- O^mn = V-j( E m — E r^ %»' (4) COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 159 where \i is the particle mass, and £ n = |^rt* are the energy levels of the particle in the well. Using the result of equation (3) for x mn ,we get _2lh_ [(-\) m - n -l)mn (5) Pmn— a (nfi — rfi) ' * ' One can easily verify that now the diagonal elements are equal to zero; this should, of course, be the case since these elements are the average value of the momentum in the corresponding stationary states which should be equal to zero, since the motion is bounded. The selection rule for the non-diagonal elements of p mn is clearly the same as for x mn . We note finally that it follows from equation (5) that p* mn = p nm , that is, that the momentum operator is, as it should be, Hermitean. The same is true for x mn . 36.* The operator corresponding to the time derivative of a quan- tity A which does not explicitly depend on the time is of the form A = \[H, A] ^1 (HA -AH). (1) The average value of A in a stationary state of the discrete spectrum with a wave function ty n , is equal to the corresponding diagonal element of the matrix of the operator (1), A = J <l>;i<l> M dx ==A nn = -jr (HA — AH) nn . (2) By virtue of the fact that the Hamiltonian is diagonal in the energy representation and that the matrix elements of A axe finite in the case of the discrete spectrum, the right hand side of equation_(2) will tend to zero identically, which proves the required relation A = . (The classical analogue of this property of A is the fact that the time aver- age of any bounded quantity A vanishes). A concrete example of this equation can be found in problem 16 section 5. 37.* The virial theorem in classical mechanics states that if the potential energy of a mechanical system is a homogeneous function of its (Cartesian) coordinates, and if the motion of the system is bounded to a finite region of space, the time averages of the kinetic energy T and the potential energy U axe connected as follows : nU = 2T. (1) The bars indicate here time averages and n is the degree of homo- geneity of the function U(x v x v . . ., x i xw), where N is the number of point particles in the system, so that we have from Eule^s 160 ANSWERS AND SOLUTIONS theorem on homogeneous functions, nU = L x ii*r ( 2 ) i = l Theorem (1) remains valid also in quantum mechanics if we understand by averaging the quantum mechanical averages (over the wave functions); we shall denote them here by < >. It is convenient for a proof to use matrix methods. (We note that a proof using the SchrSdinger equation in the coordinate repre- sentation is appreciably more cumbersome, the same applies to the proof of the sum rule in problem 68 section, 7). We have i = l Using the operator relations * _ Pi s _ M x t~Ti' Pi -~dx t we write equation (3) in the following form: 27"— ntf = 2 (iiPi+Xi'pd. (3') This operator relation is, of course, valid in any representation. We now go over to the energy representation, and use as a base the orthonormal set of eigenf unctions ty m of the Hamiltonian H=T-\-U. Since the motion is supposed to be bounded these functions refer to a discrete energy spectrum so that we are dealing with matrices all of whose matrix elements are finite - this is essential for our proof. Let us write down the diagonal matrix element (m, m)oi the op- erator (3'), or in other words, the average value of the operator (3') with respect to the wave function of the stationary state ty m . From the rules for multiplication and addition of matrices we get sn {2T—nU\ m mm (27-- nU) mm « 2 2 K^mi (Pi)im + (x&a ( Pi ) lm }. (4) We have also the well-known matrix relations ( x i)ml = -j- (Em — E l) ( x i)ml> (Pi)lm ~^^ E l — E *^ (Pi)lm> where E m and E t are energy levels of the system. The expression under the double sum sign in equation (4) is thus equal to zero and we have found the required quantum mechanical generalisation of relation (1), n(U) = 2{T). In particular, if n = 2 (harmonic oscillations) we have(t/) = (T), and if •« = -1 (Coulomb interaction) (U) = — 2 (T). COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 161 38.* We want to find a solution of the SchrSdinger equation satisfying the given boundary conditions. Separating the variables <p and /, we find a special solution of this equation (wave function of a stationary state of a rotator) of the form . torn* V.(9,0-«" f "" i '" <»-0. d=l. =b2 t ...). The general solution will be of the form Wfcp, 0-2 «»*,<*. 0. m where the constants c m must be determined from the boundary con- ditions, V (<p, 0) = 2 Cm« im * = ^ sIn * <P- m Having found the Fourier coefficients c m and having substituted them into the general solution we find the required wave function <P( ?I = A(,1— cos2<p.*~' J- (!) For values of t given by the equation ^!* — 2™ (/t = 0, 1. 2, ...), or t=*t n =*?%-n, the rotator passes again through its original state. 2n From the normalisation condition fl^Op, t) | 2 d<p = 1 , valid for all o values of t we get — ==—=. 2 YZn 39.* 1. The wave function x$( x , t) satisfies the equation /fl W = -2ir3^ (0<x<a) (1) and the boundary conditions V(0, t)=V(a, = 0, while a — a (t) =» a f(t) ( /(/) is a given function). Introducing a new variable y=s-~- equation (1) is transformed to and V(v, t) satisfies the usual boundary conditions for non-moving walls, qr(0, = ^(a . = 0- (3) 162 ANSWERS AND SOLUTIONS t Once again introducing a new variable % = J [j^jp. witn c a constant, c and putting /(f) a 9 (t), we get We have thus reduced the problem to the solution of equation (4) with a Hamiltonian which depends explicitly on the "time" * , with the normal boundary conditions (3). It is clear that it is impossible to solve equation (4) for any arbit- rary function /(*)s<p(x) . An approximate solution can be obtained from perturbation theory, if /(/) depends only weakly on the time. The given problem of slowly changing boundary conditions is thus reduced to a small time dependent perturbation of the Hamiltonian. We note that we must generalise the perturbation theory equa- tions slightly because of the non-Hermitean character of the Hamil- tonian (4'). This question of the non-Hermitean character also arises because the integral of the function W(y,t) or W(y, ■ x) depends explicitly on the time. Indeed, at any time t we have a(t) f \W(x, t)\ 2 dx = \, dx = /(f)dy, so that Oo Jl*Ttv, 0l 2 ^v = 7 ^-. (5) 2. From equations (2) and (4) it follows directly for what choice of /(O.yand t (or y and %) can be separated. This particular case arises when /(0 = /l+r-. (6) where t is an arbitrary constant. fThis choice of fit) satisfies, in particular, the initial condition /(0) = 1, or a(0) = a . ) Indeed, sub- stituting expression (6) into equation (2) and changing the zero of the time scale by taking t-\-t = t f , we get In this equation the variables can clearly be separated and we have /R . *»ty. t')=Y n (y)T n (t f ), W where nis the number of the eigenf unction. We shall not solve here the equations for Y n (y) and T n (t'). COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 163 We note only that the Y n turn out to be superpositions of Hermite functions satisfying the boundary conditions F n (0)= K n (a o ) = 0, and that the time dependence of the absolute magnitude of the T n follows easily from equations (5), (6) and (8), | T n (O | = const (O" Vl = const ( 1 + fj ' . (9) Because of the orthogonality of the Y n (y) th e ti me dependence (9) is also valid for the average absolute square V [W( a of the total wave function W(x, f), which is a superposition of partial solutions (8), oo w = l It is clear that the corresponding decrease of the probability density follows simply from the increase of the well- width using the equation J* \V(x, t) |* dx = const. ■ vt o qi v' 40.* Let us consider two inertial systems „. of reference, K(x, t) and K'(x', 0(we shall only x consider one -dimensional motion for the sake Fig. 30 of simplicity; it is clear how one can generalise the result to a three-dimensional case), which move with a relative velocity V (see Fig. 30). Let the potential energy of a particle in the field of a force act- ing upon it be U'(x', f) in the system K' . From the Galilean trans- formation x^x' + Vt' \ . t-e I < x > we find the potential energy in the system K U'{x — Vt. f)==U(x, t). <2) The Schrodinger equation for a particle of mass m in the system K' is of the form _£ W , ..^ ^^ (3) 2m dx' 2 ~*~ df' We must show that in the system K the SchrSdinger equation, 2m dx* ' ot is also valid, with U defined by equation (2) while the function W is because of its physical meaning completely analogous to W ; the quantity |*F(x, f)\ 2 ^w(x, t) is the density of the probability of find- ing the particle at time t at the point % ; this follows from the ini- tially defined connection between W and W. 164 ANSWERS AND SOLUTION. Indeed, the fact that the particle is found at a given time at a given point in space does not depend on the choice of the system of reference. We can thus put the two corresponding probabilities equal to one another, w'(x', t') = w(x, t) , (5) or, using equation (1), w'(x — Vt, t) = w(x, t). (5») It follows thus that the wave functions can differ from one another only by a phase factor of absolute magnitude unity, ¥(*, t) = e i W(x', t') = e iS i*- t W(x—Vt, t). (6) Moreover, a particle velocity v' in K' corresponds to a velocity v = v' -f- V in K , so that we get for the momenta, p — mv and p' = mv' p=*p' + mV. ( 7 ) From this connection between p and p' it follows that we must have a relation analogous to equation (5), w '(p', t') = w(p, t), ( 8 ) or, using equations (7) and (1), •w'{p—mV, f) = w(p, f). ( 8 ') We now introduce an equation for the function V(x, f) determined by equation (6). Substituting into equation (3) using equation (2) and introducing by means of equation (1) the inde- pendent variables (x, t) we get +£(3I)r-»'£-»£]*-»S- < 10 > Up to now the function S(x, t) (or rather e iS i x>t) ) played the role of an arbitrary phase factor. We choose this function now in such a way that equation (10) goes over into the Schrodinger equation (4). For this it is necessary, as can be seen by comparing equations (10) and (4) that S satisfies the equations COMMUTATION A HEISFNBERG RELATIONS. WAVE PACKETS. OPERATORS 165 These equations can easily be integrated. From equation (11) it fol- lows that mV where <p(0 is an arbitrary function of t. We determine <p (/) by substituting this expression into equation (ll f ), and we get where we have omitted an arbitrary additive constant in 5, since it would only lead to an inessential phase factor in W (see equation (6)). From equations (6) and (12) it follows finally that the wave function in K is equal to ( mV mV \ _ x „, v a *" u )W{fi—VU t). (13) As should have been expected this wave function is the product of the wave function in K' and the function (/ ip* — l ^W^ which describes the free motion of a particle together with the system K with respect to the system K. We note that the wave function we have found satisfies condition (8 f ) automatically. Indeed, the momentum probability density at time t is equal to w(/>, t) = \c(p, 0| 2 . ** e(P ' '-Tb/** t)e ~^ d »' (14) Substituting here W(x, f) from equation (13) we find easily ./»iv_pr\, ? i c( P , t) = e v-ar-»;'. i J^ ( ^ t o*" T( *"" Vi *'rf«'. — oo The integral on the right hand side is clearly (see equation (14)) c'(p — mV, t). We have thus \ 2» n ) r ' (n c(p, f) = e ^ 2« »' c'ip — mV, t). By taking the absolute square of both sides of this equation we get the required relation (8 1 ). We have thus shown the invariance of the non-relativistic Schro- dinger equation under a Galiean transformation. 41*. In the Heisenberg representation we have ±&t « -± At x(t)r=e n xe n . (1) In our case H = -p-, so that x (f) *m e W xe *vn \ 166 ANSWERS AND SOLUTIONS We shall work In the momentum representation. In that case d x = ih or dp xe and dp p* t - l -& t th 4- + £L e - dp p' 2|A» P* xe '*«' = « * 2 ^» * (i+4> Substituting this in equation (1) we get x(t)*=x + t£., which is the same relation between the operators as exists between the classical quantities. 42.* The equations of motion for the operators are of the form ■&-■£[«.* (01. dp_ dt -g-l". P<fi\. (1) The derivatives on the left are in the Heisenberg representation taken as derivatives with respect to the time coordinate which ex- plicitly enters in the Heisenberg operators. Since the operators ± — ttt e n commute with the Hamiltonian we can write equation (1) in the form, dx l I- & * . -i &t dt dt * -lit h e n [H, p]e * (2) After having evaluated the commutators we get dx _ p(t) dt f* ' ■§•-- P-'iCO. (3) Because these equations are linear they can be solved as any other set of equations for ordinary, c -number quantities. We get then } (4) p (t) «== c 2 |Ji,a) cos wt — Cjjxw sin a>t. J The constants of integration c t and c 2 axe only constants as far as the time is concerned. They can be determined from the initial conditions. Indeed, from the definition of the Heisenberg operator it follows that i At _i_ Pt = 1. <«0 COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 167 Hence we have x(f) = x cos vat -4- — sin wt, 1 ** * (5) We have thus finally p(t) = p cos <u/ — jxtojc sin atf. J These equations are valid in any Heisenberg representation. In par- ticular, for the coordinate representation we get x a (t) = x cos <*>t-\- — — sin <at -r— , Pa> (0 ~ J cos «* 17 — I 10 ** Sin "^ 43.* From the Schrodinger equation it follows that any state changes with time according to W(S, o«=r» A '«Fa 0), (1) where the set of independent variables on which the wave function of the system depends is denoted by \ . The operator A in the Heisenberg representation is of the form *&t « -^6t ( 2 ) A(f) = e n Ae n , * ' where A is the operator corresponding to the physical quantity A in the Schrodinger representation. For our problem we have AW (I, 0) = 4W(E, 0). ( 3 ) 1 fa Substituting here W($, 0) = e n W (I and operating on it from the left with e n .we get, ,-r *' Ae****? (5. *) = AW (5, 0- ( 4 ) Comparing the operator on the left hand side of this equation with expression (2) we find i(_/)W(5, t) = AW(l t). Our proof is valid for the case where the Hamiltonian H of the system does not depend explicitly on the time. To generalise it for the case of a time -dependent Hamiltonian it is necessary to replace -i At in this proof e » by the operator (t, t^ which transforms the wave function at time t into the wave function at time t. 168 ANSWERS AND SOLUTIONS 44.* The Hamiltonian of the system is of the form tf = -£+^(*-*o(0)». Our problem is to find the wave function W(x, t), satisfying the equa- tion , w AV^thS. (1) at and the boundary condition W(x, 0) = %(x). Let U(t, tj be the operator transforming W(tJ into W(0: Substituting this equation into the Schrodinger equation (1) and taking into account the fact that W(t ) can be any function, we get Taking the Hermitean conjugate of this equation, we have We now introduce the operators x and p in the Heisenberg re- presentation, x(t , t ) = UHt, t Q )xU(t, tj, p(t, t Q ) = U + (t,t Q ) P U(.t, t ). We easily get the following equations for x(t, t ) and p(t, t£ (the dot indicates differentiating with respect to t ), x(t, t Q )*=±U + [H, x]l/, P(t.t ) = iu + [H> P\V- Evaluating the commutators we find a set of equations for x (t, t ) and p(t, t ): x*=*-p, p»= — ^(.x — x (0) or x-\-n> 2 x = v> 2 x . ( 2 ) These equations have the "classical" form. Since they are linear we can solve them as if they were c -number equations, x = c x sin o) (t — t ) -h c 2 cos u> (t — t ) -f x v p = jxo)C! cos o) (t — * ) — }ao)C 2 sin u> (t — * ) + Pv where x, and p x are the solutions of the inhomogeneous equations (2) which are zero at t = t Q : x x (t, t ) = co j x it') sin o) (/ — t') dt' , p x (t, t ) = P* 2 f x o (O cos co (t — f) df. (3) COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 169 From the definition of the operators in the Heisenberg represen- tation, it follows that x(t , t )=-.x. p(t , /«,) = >. Satisfying the initial conditions we get x (/, t ) = xcosu(t — 1 ) -i- -£ si n u> (* — g + j^ (t, t Q ), (4) p(t, t ) = pcosa>(t — t ) — intixsin(ti(t — t )-+-p l (t, t ). J From these equations we easily get the solution of our problem. We write down the equation satisfied by the initial function, H W = E %. (5) Here To determine (t, t ) we can write V(f) = U(t. Q)%, or *o=-t/(0. o*(0. where W(0 is the function we want to find. Substituting this expression for % into equation (5) we get U + (0,f)H o U(0, f)W(f) = E Q W(t). <6) The Hamiltonian in this equation is the Hamiltonian H which is con- structed from our operators x(0, t), p(0, f), + (0, t)H 0(0, *)- pH °' *> .| t^^JQf . Using the relations following from equations (4) and (3), x=x(0, f) cos u)/ 4- p (0 ^ *> sin w?+ x x (t, 0), p = p(0, f)cos(ot — \m>x(0, f) sin (ot-\-p l (t, 0), or by a simple substitution, we can verify that ftffl I l*^ 2 (0. _ [p-Pjjt, 0)]2 < \u^[x — x 1 (t, 0)]a 2|* ^ 2 ty ' 2 * Equation (6) for W(t) is thus of the form [p-Pi(t,0W „ r . ^[x-xtit.0)]^ w 2(* * ^ 2 = c °* with the solution iPi (t, 0) \x-x t (t, 0)] ^(jc, t) = e * W [* — ^(f. o)]. (7) (Since in our equation t is a parameter, W can contain a phase factor which depends in an arbitrary manner on t. This will, however, not affect the required probability. ) We could have chosen a different way of determining W(t), for 170 ANSWERS AND SOLUTIONS instance, by finding the Green function of the Schrbdinger equation (compare problem 14 section, 3). We chose the method used above since the initial state was a state with a well-defined energy value. The function (7) which we obtained is the previous, zero-th order, distribution of the classical position of the particle (see equations (2) and (3)) in a system of coordinates connected to the moving par- ticles. To find the probability w n that the n -th state is excited, we must expand W(t) in terms of the eigenfunctions of the Hamiltonian at time t(t > T). These normalised eigenfunctions are of the form where n _ p //„©-(- If ^^r- (9) are the Her mite polynomials. The coefficients of the required ex- pansion are equal to oo Cn (r)= J"W(x, f)tf n (x)dx. — oo Substituting W(x, t) from expression (7) and K = V» from equations (8) and (9) and integrating over £ = (x — * ) y -^ , we find finally — OO -,-* efrpl"- »"»^t-x,)i +< Pi(*o — «i) | x /2 n nl7i L 2fi ' hi oo X J«p{-?+ 5 [/ ^p+Zf (*-*>][«- V.vnJ ^(^l — x oP I f />i(*Q — *l) I Xexpj-^ ^ ■+- 1 ft -f- COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 171 I Pj_ , ^(Xi — Xo? A 2m. Aexp and thus J ——4- — ° I Xexpi-i^ —J t £±<£lZZ*&L\ «. = i«»l , «^ r 2|x "T" 2 We note now that [|L+ ^ ( **~* &) j is the energy transferred to the classical particle which was initially at rest, during the pro- cess considered. If we denote this quantity by e, we get n n\ that is, a Poisson distribution with an average value of n equal to fi<0 We now consider limiting cases. If the point of suspension is changed fast, x{ and Pl are approx- imately equal to x i — x o ( 1 — cos ""O* Pi "= H-»*o s ' n ^ If we substitute these values into the expression for c n we get n V2*vnr°" fij Xex P\— /no) ' — 2^r+ /j V- sln ^coso)/ j. This result differs only by a general phase factor (/ -^-sino)/ . . cos iotj from the result obtained by expanding the initial function in terms of the eigenfunctions of the Hamiltonian for / > . (This expression can be obtained by putting Xl = p x = 0, in the general expression for the c n . ) If the change is slow we have in zero-th approximation *!«*<,, p x tsa , and thus W(x, f)tt%(x — x (fj). This method can be applied to solving the oscillator problem when other parameters vary with time. 172 ANSWERS AND SOLUTIONS 45.* The states of the system are described by wave functions W(0 satisfying the wave equation We have also tt?=-^P. //(0<l> n (0 = £«(04>„(0. Since the motion is bounded the spectrum of the "levels" E n (1) (2) is dis- crete (the concept of an energy "level" has here a purely formal meaning, since the energy is not conserved for a time -dependent Hamiltonian), and since we are dealing with a one -dimensional case, there is no degeneracy. This last fact means in turn that the functions <j>„(0 are real (apart from an inessential phase factor). In line with the problem we are considering we shall expand the wave function W(t) in a series of the instantaneous eigenfunctions of the Hamiltonian <|> n (0- Assuming that we know the wave function at t = , we can write its expansion for *> in the, form V-Jc.COMOexp -j\ E ^^ dt ' The coefficients c n (f) of this expansion also represent the wave function in the required representation. To write the wave equation in that representation means to find a set of equations which connect the coefficients c n and their time derivatives c n . Substituting expression (3) into equation (1) we get ^S^b-HA- -g c r$n E n) exp - E n (t')dt' tf]^ w exp —LJE n (f)dt' By virtue of equation (2) the right hand side of this equation cancels against the last term on the left hand side. Multiplying both sides from the left by <f k = ty k and integrating over the coordinates of the system q, we find, using the orthonormality of the <^> n . Ck c n exp lj ( E k ^E n)d ^j^ n dq. (4) We shall transform the last integral in that equation. First of all we have for n = k , J ^ k dq = -^-^ J tyldq = 0, since J tyldq = 1. Further, by differentiating equation (2) with respect to the time, mul- tiplying it from the left by <]>* = tyk and integrating over q, we have for J ** "IT $n d( l -f J ^k #<j>» <ty = En J Mn dc l- ( 5 ) n #= k: ^ hZn\ COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 173 Because the Hamiltonian is Hermitean and real we have, using equa- tion (2), r r » /• Substituting this result into equation (5) we get JM.*— J - e„- £ll <■***)■ < 6 > Finally, substituting expression (6) into equation (4) (we remember that the analogous term with n = k is equal to zero) and also intro- ducing the notation 4-4,-fcw Jfc(T?)t.-»-(4rL. < 7 > we find the required set of equations ^ f T?L' ° ' < 8 > n where the prime on the summation sign indicates that there is no term with n = k in the sum. The quantity/^] is according to equa- tion (7) the matrix element of the time derivative of the Hamiltonian corresponding to the transition k-+n. It is clear from our derivation that the equation (8) is com- pletely equivalent to the original wave equation (1). Let us compare the equations we have obtained with the original equations of the usual time dependent perturbation theory. Then we have H(t) =* I1 -\-V(t) where H Q is the time-independent part of the Hamiltonian with the eigenvalues £»' and the corresponding eigen- functions ty { „ (q). The latter can be chosen as the base for our repre- sentation (expansion (3) is clearly the natural generalisation of this expansion), *-2«.<0lffexp(--ff£«/). < 9 > n Substituting expression (9) into equation (1), and so on, we get for the. coefficients a n the following set of equations a * tem s2d anVkne • < 10 ) n where V*n- f tPV(0*?^. **-!#-£$?. (11) In contradistinction to the analogous equation (8) the summation in- cludes now also a term with n = k, and the frequencies of the transi- tions a> ( &£ do not depend upon the time. As with equation (8), the set of equations (10) is also completely equivalent to the original wave equation (1). It is clear that the 174 ANSWERS AND SOLUTIONS exact solution of either of these sets is in general just as difficult as the solution of the original equation (1). Equations (8) and (10) are, however, very convenient for approximate solutions if there are small factors on their right hand sides which make it possible to obtain the unknown c n or a n from their unperturbed (or initial) values, that is, from the given quantities, up to second order quantities. It is obvious that equation (10) is an adequate set of equations when the matrix elements V kn are sufficiently small, that is, when the time -dependent term V(t) of the Hamiltonian is small ("ordinary" perturbation theory), while the "adequacy" of the equations (8) lies in the smallness of the matrix elements (-— ^ , that is, in a suf- \^t Jkn ficiently slow change of the Hamiltonian //(*).• The application of the method of successive approximations in the latter case is some- times called the adiabatic perturbation theory (see next problem) which was developed in 1926-1928 by Born and Fok. 46.* We start from the exact equations (8) of the preceding problem. In general, if the Hamiltonian of the system does not de- pend on the time CM- = o) we get from equations (7) and (8) c k = const, for all ft. If the derivative !*IL is sufficiently small, though different from zero, we can put the c n in the right hand side of equa- tion (8) approximately equal to a constant, and for our problem put c n s» 8 MTO . We have thus for all kj=m t i \ m b dt' (for k = m we have, of course, to the same approximation, c m = 0). Integrating equation (1) for the given boundary conditions we get v o From the derivation it follows that this adiabatic formula is valid provided the probability amplitudes for the states which are different from the initial state are small, # „. |c*(Ql<^l for k * m ' To estimate the order of magnitude of |c*(0l it is sufficient to assume (Dfcm and (-gj-J to be approximately constant in equation (2); this equation then simplifies and we get dH 1 K (k =£ m). COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 175 On the right hand side we have the ratio of the change in the Hamil- tonian over a time of the order of magnitude of the Bohr period — — for the m -+k transition to the difference in the energy of the states m and k. From equation (3) it follows that if that ratio is small, one may apply the adiabatic perturbation theory. Thus, if over a period of the order of the Bohr period, the Ham- iltonian varies for some transition by an amount which is small com- pared to the "energy of the transition", the probability for that tran- sition will be small; this is sometimes called adiabatic invariance from the correspondence with its classical counterpart. 47.* The Hamiltonian of the oscillator is for f>0 of the form " ( '> = l^P* + y f"* a \x— a (OJ 1 . (1) where a (0 is the coordinate of the point of suspension which for our problem is equal to vj , with « = const , the velocity of the motion of the point of suspension. The "instantaneous" eigenf unctions of the Hamiltonian (1) are of the form and the matrix elements of the operator — = — fioo 9 [x — a (/)] a =» = — (1(0% [x — a (t)), evaluated with these functions, are clearly dif- ferent from zero only for the transition n = ~> n = 1 (we remember that the initial state was the ground state) and then they are equal to (m ^Ao = -^ 2T, °V^- (2) It is clear from equation (1) that the spectrum of the oscillator levels for a moving point of suspension will in general not change, that is, all <a km axe constant. Owing to this and also to the fact that the quantity (2) is constant we can completely evaluate expression (2) of the preceding problem. For the probability amplitude of the first excited state we get thus, putting coio = co, The probability to find the oscillator at time * in the first excited state is thus equal to «i(0 = k 1 (0l 8 = -^(l— cosflrf); that is, it oscillates in time. The probability that excitation has taken place during the process under consideration (that is for t> T) is equal to : 176 ANSWERS AND SOLUTIONS In order that the adiabatic approximation, which we have used, can be applied, the inequality •o> l (0<C 1 must be satisfied for all t, "o<Cl/"^. (3) Expressed differently, the adiabatic approximation can be applied provided the velocity of motion of the point of suspension is small compared to the characteristic velocity of the oscillator in its ground state. 48.* Let x t = and x 2 = a(t)be the coordinates of the walls of the well. The "instantaneous" wave functions and energy levels of the stationary states are of the form ^n = Y W) Sin ^¥)' En= ^W («= 1 ' 2 ' 3 '"')- (1) For computational reasons it is convenient to consider first of all a well of very large, but finite depth U and only afterwards go to the limit Uq -*■ oo (compare problem 25, section 1). If we take only the domi- nant terms of the expansion in —fff of the solution of the problem of a symmetric well of finite depth, we find easily t.W«/I-»[^-(2f-.)^^] <*> (for the levels we can directly use equation (1)). Let us evaluate the matrix elements of —^ corresponding to the transition n -> m. Clearly dH dU. h we have ~^r- = —^- where £7 rh = U(x — a) is the potential of the right hand wall. Furthermore, we have -^ = U h[x — a(t)}, so that the matrix elements of -^- taken between the "instantaneous" functions (2) are equal to,(e-*0), (ML>ljm] 4? m {x)¥f%(x)dx=> -^\f m (x^ n (x)Hx-a)dx - = ~U Q a^ m (a)^ n (a). (4) Putting x ■= a in equation (2) we get COMMUTATION & HEISENBERG RELATIONS. WAVE PACKETS. OPERATORS 177 which gives us after substitution into equation (4) IttL (_1) nW iM^W < 5 > We thus get for the coefficients c m of the instantaneous functions in the first approximation of the adiabatic perturbation theory (see problem 46, section 3) t i 'v <"-*>/£ nfl — /*> a Integrating we get for c m (f) , if we use the initial conditions «w(0) = 8 OT „, the expression ,, ; m = 2(-ir m + l -g±-±e ~ r i (**»)• (6) (7) o (« =£ «)• Let us now consider the problem under what conditions expression (7) leads to a transition probability from the initial state n to the m -th state which is well-defined. We note first of all that there are two possibilities: a) the motion of the wall stops at t== T; b) the motion continues for all values of t. In the first case, equation (7) is only valid for << T. For large values of t, after the wall has come to rest, the instantaneous functions are the exact ones and the coefficients e m will no longer depend on the time. Since <J>(jc. t) must be continuous at < = Twe have c m (t>T) = c m (T), (8) where c m (T) is the value of the c m obtained from equation (7) for *=» T. Since the probability of finding the particle in the n-th state is equal to \c m (t> T)\ 2 and does not change with time, there is in this case always a well-defined transition probability. This probability is equal to , ,_,„ Arfinfi f ,„ a * V ° (9) J dt li e ° Let us now consider the other case where the motion of the wall con- tinues right up to t-*oo. We shall restrict our considerations to motions where the width of the well tends to a finite limit, a (*)-*- a,*, as t-+oo- The spectrum of the system at any time t remains then discrete and the relative position of the levels does not change (no so-called crossing-over of terms). The transition probability is determined by equation (9) in which T will tend to infinity. To solve the problem of the existence of this probability we assume that the wall approaches 178 ANSWERS AND SOLUTIONS its limiting position as follows «o.-«<o~7r (t><»- < 10 > In this case we can everywhere replace a by aoofor sufficiently large values of t . We are thus led to an integral of the form oo 2_ a J f +i where <o m — " 2fi(m2 „ " a -^-. We only gave the upper limit, to stress that the expression under the integral sign is valid for t-+oo. This integral will in our case (7 > 0) always tend to an upper limit which ensures the existence of the transition probability. Let us discuss our earlier result (9) and find the conditions under which we can apply the adiabatic approximation which we have used. If T is much less than the Bohr period of the transition n-+m, that is, if ^ T< ^- h\nfl — ri*\ ' the exponent in expression (9) is clearly « 1 so that f ±Hf 1 4/l2/w2 ( Aa Y J a ai )~ infi-tmVTJ ' where Aaso(T) — a(0). The condition of applicability of the adiabatic approximation is w n ->m<C 1 that is, in our case, |M ^|«»-*»| a ^- mn Of more interest is the case of protracted action, ^ h\m* — n*\ ' In that case the exponent under the integral sign of equation (9) oscil- lates fast, so that the integral is of the order of ^~ h ■ Jjj^_ n3 . and n 2 m a /|xaa\ a /QBV *»■>» - (<B >— ih (t"J • < 9 > From equations (9) or (9") we can see that the transition probability decreases steeply with increasing \m — n\. The condition of appli- cability of the adiabatic approximation, w n + m <^l, is now of the form |„Ki!t«i. For n ~ m — 1 this last inequality means that the velocity with which the wall moves must be small com pared to the characteristic velocity 179 ANGULAR MOMENTUM. SPIN of a particle in the lowest level of the well (v^y-^^--^) (compare the preceding problem). We emphasise that the quantum number (») of the initial state is an "adiabatic invariant" when the wall moves, while the energy of the particle in that level (£„) may change very strongly (see equa- tion (1)). We note also that equation (9) is symmetric against a permutation of the initial and final states n and m, that is, it gives also the pro- bability w m + n for the transition m -*■ n. This happens because of the reversibility of quantum mechanics, that is, the symmetry of its equations with respect to a change in sign of the time. 4. ANGULAR MOMENTUM; SPIN It is well-known that the wave function transforms as follows: Y(r) = {\+id*i}*t{r). (1) under an infinitesimal rotation of the system of coordinates. Here da is a vector directed along the axis of rotation and with magnitude equal to the angle of rotation and I is the operator of the orbital angular momentum. Let us consider first of all a rotation over an angle da around the z-axis. For such a rotation we have, Y(r, 6, cp) = ^(r, 0, 9 + d a) = ^(r, 6, <p) + ^da. (2) Comparing equations (2) and (1) we find, To find the form of the operator l x in spherical coordinates we perform a rotation around the x-axis. We have then f(r, 6, ,)_«,. a + de, ,+*)-{> +(£|+££Htfr. 6. ,), from which it follows that Let us evaluate ^L and ^1 da da ' One sees easily that z 1 — z = — yda, y' — y = zda, 180 ANSWERS AND SOLUTIONS and since z' = r cos 8, y' = r sin 8 sin <p we have, di and thus ^L^ — siiKp, ^ = — ctgO-cos* da. ' "« . / aJ ==/(sin ? ^ + ctg0.cosc P ^). Similarly we find / 5. y== _^cos ? ^ — ctgO-sincpA) \ z , = /^ cos ixz') + 4 cos Cv*') H~ 4 cos (^ ')• 7. The transformation can be written in matrix form: e<(<H<p)cos 2 — , -4= e*? sin 6, — g-^-'pJsin 2 — 2 ]/2 2 -4=^ sin6, cos 8, -Le-^sinB, — *«*-*) sin 2 -, —-e-* sin8 , .*-«++*) cos 2 - 2 »/2 2 Yu Yu-i Y[ Q 8. Using the result of the preceding problem we find for the case M = 1 , 1 . w{-\- l) = cos 4 -^-, w (0) = -^- sin 2 6, w{ — l) = sin*-~-, for M = 0, w(+l) = lsin 2 e, w(0) = cos 2 6, w(— l) = ysin 2 8 1 . and finally for M = — 1, w(+l) = sin*-|, w(0) = ysin 2 8, w(+ l) = cos*-J 9. W (+I) = cos 2 l, w(-±) = tin*±. The average value of the spin component is equal to ycose. 10. We use the matrix of the transformation of the components of the spin function under a rotation of the coordinate axes. This matrix is of the form / 4-0M-+> 8 . 4-(<p-+) . 6 \ e a cos ~o le '' sin "2 \ie 2 sin-^- e 2 cos Using this matrix we find the spin function in the new system of co- ordinates *-(<p-«W+*P „._jj_ T — £(<P+-V)+iP - Q rr»c 2 ty\ = e 2 vr ' T " " cos 4 • cos8-|-te 2 " T "" r sin— -sin 8, ^,= ie " sin -^- • cos 8 -)- e 2 " "' cos y • sin 8. ANGULAR MOMENTUM. SPIN 181 We find the probability that the spin is directed along the z'-axis, w i = tf'i'i = cos2 2 " cos2 S "f" si " 2 o" * si " 2 8 +~ + T sin sin 28- sin ($-{- a, — (3). From this formula it follows that the probability for the value of a spin component along the orbital direction depends only on the dif - ference a— 3 and is independent of a and /3 separately. 11. The spin direction is determined by the angles 8 = 28, $ = £-f-p — a. 12. It is possible. In the case of a mixed ensemble for every direction of the inhomogeneous magnetic field one will always get a splitting into two beams. In the case of a pure ensemble by a suitable alignment of the instrument one can obtain the disappearance of one of the beams. 16. For an infinitesimal rotation over an angle da around the jc-axisthe components of the spin function are changed as follows: (l + /*faO| to where Expression (1) is equivalent to three differential equations (1) da /J 70 ' d« _ "-i/T(to-+-'!'_i). to. Y2" da Yl By solving these equations one obtains easily the transformation matrix in the following form: cos*- J- 2 V2 V2 — sin 2 — [ 2 sin a • 9 a ■ sin 2 — 2 sin a cos a Y2 sin a — =sin a Y2 9 a cos 2 — Similarly one obtains the transformation matrix for a rotation over an angle a with respect to the z-axis, 182 ANSWERS AND SOLUTIONS /e ia \ J. \0 «-*■/ We shall use the Euler angles y, b, cp to characterise the rotation. We must then multiply three matrices to find the required trans- formation matrix. The final result is £<(* + ?) cos 2 — 2 — -js e*+ sin 8 2 — = e*? sin 6 Y2 cos 6 Y2 «-**sin6 -e-i(-V-<p)sin 2 — 2 -i=e-*-Vsin6 g-i(*+?) C OS 2 -5- 2 (2) We note that we could have obtained the same result by considering the transformation of a symmetric spinor of the second rank. Bet- ween the components of a spinor function and the components of a symmetric spinor there exists the following relation: il,21 = . ■■%> Y ,1.22 — d Y-f (3) A spinor of the second rank transforms as the product of two spinor s of the first rank, that is yii = a2( j,n _f- 2apt|> 12 + p 2 y 22 , Y 12 = a-r^ 11 + (a8 + p T ) <> 12 + m 22 > ,1/22 = T 2,i,ii _j_ 2 T 8-> 12 + 8 2 y 22 . If we now replace the components of the spinor by the components of a spin function we get by applying equation (3) 44 = 0^^4-/20^4- pVi. 4£ = V 2a T?1 + ■ («» + PT) 7o + V 2P^_ V If we then substitute in this relation the values of the coefficients t(i»++> *2 cos -jp p = /siriy -tCp-V) -5-<*-+) = e — S-(?+« cos 2-, 7 = i!sin-2- we find expression (2). 17. To find the required probability we use a formal method which consists in considering instead of a particle with angular mom entum / a system consisting of 2j particles of spin £. Since in our problem the angular momentum component of the particle is equal to j, all particles in the equivalent system of 2j particles must ANGULAR MOMENTUM. SPIN 183 have a z -component of the spin equal to +£. The probability for a spin component +^ (or -±) along the z-axis is for each of the par- ticles equal to cos* 1 (or sin* -| ) (problem 9 , section 4). In order that the value of the z -component of the total angular momentum of these particles is equal to m it is necessary that y'-J-m particles have a ^-component +i and the remaining j—m particles a «-compon- ent -£. The required probability w(m) is obtained by multiplying (cos'!) X(sin^) by the number of all ways of distributing 2 j particles into two such groups, that is a + < ff^_ w)! • In this way we get "(»)- (/ + .)!»—)■ ( C0S 2j (—7) • One verifies easily that ^w(m)=l. -i 18. The states of a system with angular momentum / can be j . -j+m ' . j c J-m'- — h described by a symmetric spinor \i 1 1 1 1 1 1 1 1 1 11 11 i\i 1 1 11 iii\2zzzzzz\.2Ziz\ of rank 2 / . To solve the present [ j ! j problem we must establish a con- | ill! nection between the components \i 1 1 1 1 1 1 1 1 1 1 1 ii\zz zzz\z zzzzzz\iiuif\ j+m^ j-m j + m> j-w j— j+m-v -i* J-M .-4— v— j <j,n "Tm 2Frr~2 j j,'iHt7i srTrrs Flg " 31 From Fig. 31 one sees easily that J+M' J-M' *'uT7i 2lt~2 = (J+M')ijj—M'y v T (27)! A x y (27)! - (Tnp^-'+^+'-w-*^ -^ |— 2 X ^ t/,/J vl(Af^-Af + v)l(y + A|-v>|(y-AI^->)l ^ " "'' where a, 0, y, and 6 are the Cayley-Klein parameters. Since J+M J-M ~n = y v+Mn;-M V HM}< J+M' J-M' and t|» (M) = 1 we have <K (AT = W-h MO! (J— M')\ (/+ M)\ (7— M)\ X y ( T ) v (p)M'-M+v (g) J-+3f-v (5) J--^-v A . ^ v! (Af _ Af -j- V )! (7 + Af — v)! (7 — AT — v)! v = 184 ANSWERS AND SOLUTIONS It follows then that: P(M, M') = (J-\-M')\(J— M')l{J+JVI)\(J— M)\{cos^j X x v ( -' )V ('^) . . A .£v!(Al' — M4-v)!(/ + M — v)!(/ — AT — v)! j In the summation we must put all terms which lead to a negative number in the factorials equal to zero, that is, in other words, we must sum over v in such a way that v satisfies the inequalities v>Al — AT, v<7— M'. 20. We find first of all the eigenfunctions of the operator j z . To do this we write the operator j s in matrix form /.= since {, = — 1 4- . the equation determining the eigenfunctions and eigen values of /, is of the form or _/*b_i It follows that where / x and / 2 are arbitrary functions of r and * and m is a half integer. From all possible functions of the form V,(r.»).'<" + r>V we must select those which are at the same time eigenfunctions of the operator P. Such eigenfunctions are of the form ANGULAR MOMENTUM. SPIN 185 The last stage in this construction will be to make this function an eigenfunction of the operator of the square of the total angular mom- entum by suitably choosing # t and R 2 . To do this we write the equa- tion /2<j» = / (/ 4. !),<,- in matrix form /„ //. \fRx(r)Y lim _, (ft, <p)\ and take into account the properties of the spherical harmonics (4 + ily) Y im = V(l-\-m-\-l)(l — m) Y lf ^ (4 — ily) Y lm = Vi.1 — m+\){l-\-m) Y l% m _ v It follows then from the matrix relation which we have written down that /?! and R 2 must satisfy two homogeneous equations [/(/+i)-y(y-hi)+/n + |]^ + |/"(/+j) 2 -/« 2 /? 2 =o > t^( / +4-) 2 - w2/? i+[ / ( / + i )-^o'+ i )-^+4] /? 2=o- In order that these equations can be solved, it is necessary that j is equal to either / + i or I- \. If we put j = I + \ we get and thus *i = "/"'+4+ "KM. /?.= ]/"/ — w + y/?(r) ^(/.7 = /+y /«)=/?(/■) r 2/-fl r i-m-y, f 2/+1 'J-TO+V* (/=o, i, 2, ...); similarly we get for j = I- \\ Cf(l,j = l — ^, ^ = /?(r) 2/ + 1 2/+1 J, Wl-l/ 2 J. W + l/l (/=!, 2 3, ...)• 186 The factor 21. 1 V2/-+T ANSWERS AND SOLUTIONS follows from the normalisation. orbital ang.mom. m — Ya spin Vi PROBABILITY j = l + l k J = l- l l% / + «+! 2/+1 2/+1 orbital ang.mom. ™ + V 2 spin -Va 2/ + 1 /-fm+1 2/+1 / z a=/+v 2 ): 2wz/ 2/ + r * z (;=/+7 2 ) 2/ + r W = / — V,)^ 2 ^^ ., s,w=i— 7 2 ) = 2/+ r 22. The eigenf unctions of the operator of the spin component in the direction 8, $ can be found from the relation /a\ /a (a^ sin 8 cos $ -\- c y sin 8 sin $ -)- c z cos 8) ( fi ) = from which it follows that a sin 8^*— p cos 8 = ,3. From the last equation we find the ratio —■ (1) (2) On the other hand from the explicit form of ,the functions Cf(l,j = l-\--^ t m\ and <M/, y' = / — -^, rnjyre find where = £ . - : = £ . __! g~ l 9 -■ /"' + "» + Va -: • ,11/ for j = l — x \i. (3) ANGULAR MOMENTUM. SPIN 187 If we compare equations (2) and (3) we find that |= 9, , that is the spin direction in the given point of space lies in the plane through the 2[~axis and the given point. The angle & is determined from the condition e p;"-^(cos») 23. The operator of the square of the total spin is equal to *=* + *+ 2^ = |(J ?X(o ?),+ -r?{(i 0U1 o) 2 +(/ "o'lC oOi+lo -1)1(0 -l),}' where the indices 1 and 2 indicate the number of the particles. We determine the eigenfunctions of this operator first of all in the case where the components of the total spin are equal to zero, *MU(U+»(!),(;)J=M'(J)A+»(!),(J)J- We find then, (X— \)a — b = 0, — a + (X— l)£c=»0. For X we have two values, X= 2 and X = 0. For X = 2, a = b and for X = a = — b. If we take into account the normalisation condition c 2 -{-b 2 = 1 we find the eigenfunctions in the following form : *KJ)A+(a(l)J x - 2 (S=1) ' One can easily verify that the functions ( ^ ( j^X 1)1(1)2 are also eigenfunctions of the operator S 2 , where in the first case the 2-com- ponent of the spin is equal to 1 and in the second case equals - 1. The functions we have found are also eigenfunctions of the operator 24. The wave function of the system T(J, M) has the form of a sum of products of functions of separate particles, where the lower index on the wave functions indicates the value of the angular momentum component. The coefficients c t must be determined from the condition J*W (J, M) = J (J + 1 ) V (J, M). (1) 188 ANSWERS AND SOLUTIONS The wave functions of the first particle can be written conveniently in the form !/») = ?i <LU) = TO If we use that notation for the wave functions, the operator / will be a 3 x 3 matrix /0 1 0\ t<™ — I V2 1 ^0 1 0/ - ( l ° /,.= o \0 and we find for the operator J 2 : >=/?+/*+ 2^4= //(/+ 0+2 + 24, /24_ = ( V2/ 2+ /(/+i)+2 V V2l 2+ where 4 = 4+'4» 4=4— '4- If we use the properties of the operators 4 and I _ , M"m = / (/ + m+l)(/ — m) <K, +1 , we find that condition (1) leads to two equations, [7(7+1) — /(/+!) — 2M)c l =V2 V(l + M)(l — M+-l) c Q , [7(7+ l)-/(/+ l)+2M]c_ 1 = /2 /(Z + Af + i)(/_») Co (the third equation is satisfied identically). If we solve these equations we find for c (7, M) : / Cl (/+1, A1) c (/ + l, M) c.jCZ+l, M)\ I <?!(/, M) c (/, Al) c_ x (l, M) ) = Vc^Z— 1, M) c (/— 1, M) c_ x {l—\, M)l .yi + M)(l + M+1) 2(2/+l)(/+l) + M){l-M + 1) ^ + M + 1)(<- (« + !)« + 2/(1+1) Kza+i) m + d l/"(£ 1) r •M)(/-Af + 1) 2(2i + l)(Z+l) + M + l)(l-M) 21 (1 + 1) */~ {l-M)(l-M+l) y/- (lJ r M)(l-M) l/"(£ V 2/(2/+l) r /(2/ + 1) V + M)(l+M + \) 21(21+1) ANGULAR MOMENTUM. SPIN 189 Since this matrix is orthogonal, its inverse is the same as its transposed matrix and therefore each of the functions tf ^_ u $y&, 4»-W+i can be expressed as a linear combination of *"(/-}- 1, M), W{1, M), *P(l — 1, M) with coefficients which are in the columns of this matrix. 29 « h, = ?, = 0, 30. - 0.24. 31. - 1.91. 32. a) 0.879; b) 0.5; c) 0.689; d) 0.310. 33. The weight of the D wave is equal to 0.04. 34. a) — j^r 2 ; b) go r*. One must take into account the dif- ference in sign of the quadrupole moment in the i P 1 and 8 Pi states. 39. The states with well determined values of J (J 2 is an integral of motion) can be constructed from the states with L = j—i/ 2 , ■£ = y+V»- An inversion (*-*— x, y-> — y, and z-+ — z) will not change the Hamiltonian operator of the total system (parity is an integral of motion). The parity is different in the states with L— J— V» and L = J-{- 72 • It follows from this that in a state with given value of J the orbital angular momentum I corresponding to the relative motion of the particles must have a completely well- determined value. 40. The spin of the a -particle and the spin of the nucleus B axe equal to zero so that the orbital angular momentum, L , corres- ponding to the relative motion of the a -particle and the product nucleus is equal to unity. It follows that this system is in a state of odd parity (the a -particle has even parity) if the original nucleus had even parity. 42. It is impossible. 44.* The required operator (which we shall denote by R?,) must transform any arbitrary function of the coordinates of the system W(r v r 2 r i r N ) into the same function W but of coordinates which are rotated over the given angle. For a rotation over an infinitesimal angle d<p around the direc- tion n (where n is a unit vector) the radius vector of the /-th particle r t will be increased by dr i where dr 4 = d<? (» X r$. 190 ANSWERS AND SOLUTIONS Under a rotation over a finite angle f d<? = <p around the same axis n /*; W iH undergo a finite change jdri = f dcp(» X rj = n X / M<P s ^. (1) Let us consider the most general operator R which is the opera- tor of arbitrary finite displacements of the particles in the system. (Strictly speaking these displacements must satisfy the condition div &r = 0). To determine R we have RVirt r., .... r N ) = = ¥(^-1-5^ r. + hr v .... />+8/v). < 2 > We shall expand the right hand side of equation (2) in a Taylor series; this expansion can be written in the form W^-fo/v ..., r. + Sr. ^ + 8^) = t=l +i(i".^) , *+...-[i+i«r,^+ \ j=i / L i=i +^(1^)'+-]^. r * ^ Comparing equations (3) and (2) we can convince ourselves that the expression within the square brackets is the operator R. Since that expression is the power series expansion of the ex- ponential function we can write R in the following symbolic form (we introduce the operators of the momenta of the particles p t = —jf~) In the particular case of a parallel translation of all particles over a distance a we have 8r t = 3r 2 = . . . = 8r 4 = . . . == tr N = a, and equation (4) will be of the form Ra ^^ I a == # » (3) ANGULAR MOMENTUM. SPIN 191 N- where P = 2 A is tne operator of the total momentum of the system of particles. We can now go over to the required case of a rotation over a finite angle. If we take the »-axis along the 2 -axis of a system of cylind- rical coordinates (p, 9, z) and substitute l< ?i = cp , lp t = lz t = 0, we have Substituting this into equation (4) and bearing in mind that — is the operator of th< of the i-th particle we have 2L JL is the operator of the z -component of the angular momentum l Oft R , . '=« , (5) <1u N where M^ V-yiT~ is the °P erator of the component of the angu- lar momentum of the system of particles along the axis of rotation. Finally we have in vector form (M g z=z M n = nM)'. Rf , n — e 45.* Because of the equivalence of the x, y, and z -axes we have clearly M 2 == Ml + fii;-\- M; = M% -\- M; + Mt = 3M' x . From the definition of the average value of M x and the fact that all different possible values are equally probable we get 1 1 _ 2 ma 2 2 mS ,,2 *2 2 t 2 m = -l tA »« = ! M^fim =fi -27+T =fi 27+T"- Using the well-known equation for the sum of the squares of integers, JL m - § » 192 ANSWERS AND SOLUTIONS we get and thus, m x = g , 46.* It can be shown that momentum and angular momentum of a system are simply connected with the operators of infinitesimal translations and infinitesimal rotations of the system (more exactly, they are proportional to the difference of those operators and the unit operator). Any translation commutes with any other translation so that the operators of the different components of the momentum will also commute. Two rotations around two non-parallel axes will, however, not commute with one another which corresponds to the non-commuta- bility of the operators of the different components of the angular momentum. 47.* Since for any of the particles there are possible (2/-f-l) spin orientations, the total number of independent spin functions for the system of the two particles A and B is equal to (27-f-l) 2 . These functions, which are not symmetrised with respect to the spins, have the form x^X* 3 '* where /, k== — /, — 7-f-l, •••» / — 1. /• If we symmetrise them we get the following result: for / = k there are (2/-h 1) functions of the form x[^xf } (symmetrical); for i =£ k there are (2/+ 1)2 — (2/-|- 1) = 27(27+ 1) functions of which one half, that is, 7(27+ 1) functions are of the form x ¥*)$* + Y* } X ( f ' (symmet- rical) and 7(27-)-l) functions of the antisymmetrical form x^X* 8 ' — — xW '• In this way we get altogether (7+ 1)(27+ 1) functions which are symmetric with respect to permutation of the spins of the par- ticles A and B and 7(27+ 1) antisymmetric functions which lead to the required ratio -y-. 48.* Let us denote the totality of the coordinates of the /<-th particle including, for particles with non-zero spin, the spin com- ponents by 5 i (/= l, 2, 3) and the total set of quantum numbers which characterise the ft-th of the given "single particle" states (k = 1, 2, 3) b y Pk- This set consists of three or four quantum numbers depending on whether or not the particles have spin. Since we have assumed the interaction between the particles (we consider only ordinary and not exchange interactions) to be small, the required wave function of the system will be a linear combination ANGULAR MOMENTUM. SPIN 193 of products of "single particle" wave functions <|> p (^). . Since we considered Bose particles the total wave function W(Z lt £ 2 , £ 3 ) must be symmetrical with respect to the permutation of any two particles, that is, the linear combination must be the sum of products of the form with all possible permutations of unequal indices p k . We will assume that the functions <|> p (^) are mutually orthogonal and normalised, JtifcCtoVW***- 8 *- (1) By integrating over d\ we understand in general both integration over the coordinates and summation over the spins; 8 W is a pro- duct of the corresponding number (three or four) of 8-functions. The normalisation of the total wave function will be different depending on how many of the constant p v p 2 , and p z are different. We must consider three cases: 1) All three states are different, that is, symbolically, Pi=t*Pz* P 3 - The normalised wave function of the system is given by -h <fo &) *A (&2) 4>P, (S3) + fc. (Si) *A (*2) *A (S 3 ) + H- **. (Si) 4>p, (S 2 ) 4>*, (Ss) + **. (Si) *a (S2) **. (S3) + < z ' + <MW <!>*««) <M$8)J- Indeed, integrating HFfo, £ 2 , 5,) | 2 over the variables l v i 2 , ^ , all cross terms of the form ff I J {^(Sl)^,(S2)^,(S3)^(Sl)^ 3 (S2)^(S3)^ 1 dS 2 rf$3 are equal to zero because of relation (1) so that from the general number 6.6 = 36 of terms which enter into |W| 2 there are only six integrals left which are the integrals of the absolute squares of each of the terms in the sum (2), S J J Jl^(Si)l 2 l^.(S2)l 2 l^ 3 (S3)NS 1 rfS 2 dS 3 . Each of these six terms is equal to % by virtue of relation (1) so that, indeed, - - - „ J J ]>,..*..*, (Si. S 2 , S 3 )| 2 *i*MS 3 =l. 194 ANSWERS AND SOLUTIONS 2) Two of the three states are the same, for instance Pi •£ Pi = Pz- The normalised wave function of our three Bose particles is equal to -h to, (Si) to, (5 2 ) to, &) + to, (^i) to, &) to, (6s)} • Indeed, in the same way as in case (1), of the nine terms in the normalising integral j J* J* | W | 2 d^ dl 2 d% 3 only three are different from zero and each of those is equal to (y gr) • ! = y so that the total integral is equal to unity. 3) All three particles are in the same state, that is, A = Pi = Ps- The normalised wave function of the system is clearly ^fc A (?i. * t . y = to 1 (Uto 1 (yto,(y- 5. CENTRAL FIELD OF FORCE 1. Putting Rni^— 1 the equation for R n i goes over into — 2JT Xnl — [ E nl —UV) 2JX72 ■ J Inl ~ «• This equation is formally the same as the Schrodinger equation for a one -dimensional motion in the region < r < oo with an effective potential Since xm = rR nl tends to zero for r = , we can assume that U = -f oo for r < for this one-dimensional problem. 2. In the equation for x = R r *r + $lB-"W-!2£2} x -o CENTRAL FIELD OF FORCE 195 we make the substitution where A and 5 are real functions. If we put the real and imaginary part of the equation separately equal to zero we have 2A'S' + S"A = 0, (1) From the first equation we get . const A= . . The second equation we can solve approximately, assuming h 2 to be a small quantity; it is necessary, however, to bear in mind that when we make the transition to classical mechanics (h -► 0) one must assume that hi is finite since hi is the angular momentum in classi- cal mechanics. We can thus in equation (2) only consider the term —j- to be small. For small values of r , when the dominating term on the right hand side of equation (2) is m V + l ) we have s ,_ ihYi(i + \) f A _y- SQ ^ we f . nd approximately ^^__ Jl . We thus obtain a better approximation for S if we take this term into account by substituting this approximate equality into equation (2) (such a correction does not exist for large values of r ). We find thus s= J/^iE-i/^-eff+ia,,,, A = const 3. We shall write the Hamiltonian operator in the following form, // = // o + ^-7H'Where//o = -2^ 5 - r (/ ' 2 Tr ) + U(r). The smallest values of the energy and the corresponding eigen- functions are related by the equations 196 ANSWERS AND SOLUTIONS The last expression can be written in the form Let us compare the first term in this expression with E? ia . Since corre /(/ + i) A 2 fy corresponds to the minimum eigen value of the operator tf -f ^ x ^ we have As far as the integral J- {i ~^ l) ?i+i?m^ is concerned it is always larger than zero. Hence Ef a < E?S, that is, we have proved the statement given above. 4. p x + p 2 =P = — tt V B ; / t + ia^Z = [/?/>! + [rp] , where /> = — iftV r . 5. The potential energy is U (r) = ^^— . The radial part R of the wave function satisfies the equation If we substitute x = Rr and use the notation /l /i we have ■XT 4- (* 2 - *r* - 1 -^±±] 1 = 0. (1) If we consider the asymptotic behaviour of x as r -» and as /■ -> oo, we can put the solution for x in the form CENTRAL HELD OF FORCE I97 If we substitute expression (2) into equation (1) we get an equation to determine the function u(r) tt "+2(^±i — \r)u'— {2X(/ + 8/ 2 )_ iktj Bs=0> (3) By introducing a new independent variable \ = Xr 2 equation (3) goes over into the following differential equation where 2\ ~~ fiw ' The solution of this equation is the confluent hypergeometric function « = F{i-(/+|~ s), /+»/,; l}. Requiring that R vanishes as r -> 00 we get "H'+T — S )= — n r (»r = 0, 1, 2,...) and hence we find that the energy levels are equal to E n i= MH- -f" 2/i r+ 8 /2)' and the wave functions given by ^n r im=^ rt F{-n r , /4- 3 / 2 . ^ 2 }^,„0>, «p). 6. The wave functions are ®n in ,n,(x, V, 2) = <p Bl (*) <p„,(.y) ?„,(*), where ?«(*) = -7= - -; XX — jj-) e V2 n \ n+l nl fa V djr/ The corresponding energy levels are (see problem 5, section 1) En ini n 3 = hm (»! 4- n 2 -+- n 3 -+- -|) . The connection between ^ M lm and ^n,^ for « r = 0, l=\ is of the form : 198 ANSWERS AND SOLUTIONS +on = 7f<*ioo+'*oio). %10 = *ooi« ^oi.-i = ^( $ ioo-'®oio)- 7. Z B = (« + !)(«-+- 2), re 8. For |He we have where r ' P( r ) = 7— ^^ 1 / r \ a e 2 (r V"27t) 3 where '.-/pi R=r »- For ^O we have h\,-t(t7)\ pO= s?( 1 + 7) (r /2t /? = 3.73r . 9. The equation for the radial function is of the form where p is the reduced mass, p = ^ ^^ « T , since M p « M„ = M. If we put / = and # =^P-, we find Introducing a new variable *• we find &L+1.*L-L.(c*—£)y=0, "Sp + e rfM V &) x central Field of force 199 where k* = — ^Ea*>0. The general solution of this equation is X = BMct) + B 2 J_ k (rt). As r -► 00 (S = 0) the wave function of a stationary state must tend to zero so that B 2 = and hence In order that R is finite for r = we must have J k (c) = 0. This equation gives us the connection between a and A. To obtain values for a and 4 referring to the ground state, it is neces- sary that c is the first root of the Bessel function (the radial wave function must not have any nodes) : a • 10 13 k c A t MeV 1 0.45 3.1 100 2 0.91 3.7 36 4-4 2.02 5U 14 10. The average value of the energy E in the state described by the wave function <]>(/•), is given by the following expression: According to the variational principle the quantity E takes on the value of the ground state energy if is the exact ground state func- tion. If we take for a function depending on one or more para- meters a, 3, . . ., the energy E will be a function of those parameters, E (a, p, . . .) and the best approximation to the energy and the wave function of the ground state will be obtained for the values a = a , p = 3 , . . . , which satisfy the conditions 200 ANSWERS AND SOLUTIONS The quantity E(a , p Jf . . .) is always larger than the ground state energy and is the nearer to it the larger and more expedient the class of trial functions taken. l — In our case <b = —== R (r), where R(r) = ce 2a . From the normalisa- tion condition it follows that c 2 = ^ so that CO CO /. , or /» ar r E(ol) = c 2 ~ J (£) e u r 2 dr — c 2 A\ e a a r 2 dr = o 6 _ A^ /_a_\2 _ , / a \3 We find the minimum of E (a) rfa ~" 4(JLa2 (« + l) 4 " Hence, (« tt +l)l =a Jg^ = 22 . 3; ao== l.34. The value of the energy for this value of the parameter is £ = — 2.14 MeV. The exact solution of this problem gives for the given values of A and a the value E = -2.2 MeV. (See preceding problem). 11. The equation for the radial part of the wave function for r < a is of the form ^§)-'-^«+^ = ' « where R ~ m ' while for r — a, R = Q. We introduce instead of R a new function % ( r ) through the formula v> Substituting this into equation (1) we get for x( r ) tne equation the solution of which are Bessel functions of half integral order, CENTRAL FIELD OF FORCE 201 R(r) = ^=J l+1 , t (kr). Y~r The value of the energy E = -L- of the stationary states is obtained by requiring that the Bessel functions tend to zero at r = a J, +Vl (£a) = 0, while c follows from the normalisation condition. The simplest energy levels to determine are those for particles with angular momentum / = 0. In that case 4**0 = /5b si and the energy is given by sin&r 12. The problem reduces to the solution of the one -dimensional problem with potential (—U 0<r<a, "(') = ] r>a, I oo r<0. If we put in problem (4) of section l u 1 = oo, U 2 = U , we get the equation which determines the energy levels of the discrete spectrum ka = ntz — arc sin hk k = Y2mB Fig. 32 YZrnUo These energy levels can easily be obtained using graphical methods (see fig. 32). The depth of the well for which the first dis- crete level occurs is equal to U ° ~ 8ma* • 13. When the well edge is rounded off all levels are changed upwards, that is,. A£ > 0. States with large values of 1 will suffer large shifts of the levels, since the particles in a state with a large value of the angular moment spent a relatively large part of their time near the edge of the well. 202 ANSWERS AND SOLUTIONS 14. The radial wave function satisfies the equation In the region I, where U = 0, the solution which tends to zero for r = is 1 = A sin &r, « = -p" • In the region II (U = U ) the general solution is of the form y = B + e "-\-B_e , / — p The coefficients B + and f?_ are determined from the condition of continuity of »x and x' at the boundary of the regions I and II, Aslakr l =B + -\-B_, Ak cos kr t = x (B + — B_). Hence R = 4-( sin kr,-\- — cos kri\, (1) B_ = y (sin kr x — -£ cos kr ) ' The solution in region III, where again U = , is The condition of continuity on the boundary between regions II and III leads to: B +e x{r '- ri) -\-B_e- xlr '- ri) = C + -\-C_, x (B + e x <'•-"•> - B_<T * ( ' a - r,) ) = /* (C + - C_). We find thus c + -!b + (i+£).-«-*'+±B-(i-£).""^. C- = T B + ( 1 -/T)^ ( "" r ''+i s -( 1 +s)^ ,r -"' Using equation (1) we can express C + and C_ in terms of A , C + -|^-«»r 1 (n-j).^-' J jl+ I -^f.-"^- , + CENTRAL FIELD OF FORCE 203 -h-ctg^i 1-4 1 + Ik (2) (2) Expressions (1) and (2) determine the form of the stationary wave functions of the particle. The behaviour of the wave function depends essentially on the particle energy. Let us consider the dependence of C + and C_ on the energy. We will assume that x(r 2 — r t )^> 1. We can then neglect all terms which contain a factor e~ 2x(r *~ ri \ and we have C + »4sIn*r 1 (l + *)e-<^>{l+A c tg* ri j > C _=C\. If the quantity within the braces is not too small the coefficients C + and C_ are appreciably larger than A, that is the wave function is appreciably different from zero only in region III (fig. 33a). For some values of the energy when the expression within braces in equation (2) is small C + and C_ can be anomalously small. Such energies will be in the neighbourhood of the values E n which satisfy the transcendental equation W-u^, •Ctg /"*£«'. = <> and which are the so-called quasi- stationary levels . % Fig. 33 204 ANSWERS AND SOLUTIONS One can easily show that the values E n are the true discrete energy levels of the problem with a potential depicted in figure 34 (r 2 -► oo). {UfD U„ Fig. 34 The values of the energy which are lying in a narrow band near the quasi-level correspond thus to wave functions which are vanish- ingly small in region III (fig. 33b). The probability to find particles with a strictly defined energy within region I is equal to zero. Indeed, the wave function of a par- ticle with a well-defined energy belongs to the continuous spectrum and the integral over region III of | <|» (r, E) | 2 diverges; at the same time the integral over region I is finite. This statement remains valid even for states near the quasi-level. To find, therefore, the probability of emergence of the particle from region I it is necessary to consider states which are a super -position of a number of station- ary states with nearly the same energy, that is a "wave packet" whic is localised in region I, and to investigate its "spreading" with time. We shall take for the wave function at t = a function Xo which is practically equal to zero in region III and which in the regions I and II is the same as the wave function of the quasi-stationary state. We expand XoOO * n terms of stationary wave functions, Xo(r)= f <?{E) tE (r)dE. (3) o The functions i E {r) will be supposed to be normalised in the energy scale. The state of the particle at time t will be Xo 0\ t) = / cp (E)xe (r) e~ % ~* % dE. We consider the probability that the particle during a period t will have gone into the initial state Xo (O , l—z- t ^(0 = |/xo(OXo('-. 0d/-l 2 = |/l<P(£)1 2 '" lft d E (4) CENTRAL- FIELD OF FORCE 205 The problem is thus reduced to finding the distribution in energy of the initial state, |cp(E)| 2 . From equation (3) it follows that oo <f{E)=fxo(r)XEi. r ) dr - (5) o In agreement with what we have said, one can take for Xo( r ) an eigenfunction of the auxiliary problem with the potential depicted in Fig. 34 for r<r ly Xo (r) = asin& r ^ = ^) and for r > r t The value of k Q is determined by the condition sin k a r, = — — , cos k n r. = — , and the normalisation constant is given by a = y y 2% The functions % E ( r ) were in regions I, II, III determined only up to a general multiplying constant A. We must now choose A in such a way that the Xs( r ) are normalised in the energy scale. The asymp- totic form of x^(r) is determined by the values of the coefficients C + and C_. The normalisation oo fx E (r)-&,(r)dr^HE-E') o leads to , /-•— We can thus determine the dependence of A on energy using , |C ± | TIT 2^ equation (2). We noted earlier that the ratio -j^ = A{E) was large for practically all values of the energy and small only when E is near one of the quasi -levels. The function cp(E) is thus a steep h 2 k z maximum near E = -^ . In the region of the other quasi-levels because of the nearly total orthogonality of the function XoC'O *> tne 206 ANSWERS AND SOLUTIONS eigenfunctions x E ( r )' which refer to the other quasi-levels, the inte- gral in equation (5) will be nearly equal to zero, even though A(E) will again increase. In expression (4) for the probability W(t) only the region of values E which are near to E Q will give an appreciable contribution. After these preliminary considerations we can attack again the evaluation of cp (E). First of all we find the dependence of A on E. From equations (2) for C + and C_ it follows that : Further, B + and B_ can be expressed in terms of A : B + = 4 ( sin kr i + T cos kr i) ' B _ = ^ (sin kr x — -£- cos kr^ . We can put k — k = Mi near the quasi -level and assume that the following inequalities are satisfied, |Afe|<C*oand|Afc|<Cx. W The dominant terms in B + and B_ will then be B = — A*-. Assuming that r ,(rr,,) <l, we have and since l c +l =-g" K 2& ' we get *2 e -2x(r,-r,)\ 2 1+Wl 4£%2 X ( rj _ ri ) i4(£) = 2 (l + wi) V A < i *> , +(f•TT^- !,,r -" , ) 2 We can now easily evaluate the integral (5) which determines 9(£) if we assume that as before the inequalities (6) are satisfied. A (E) The function x E ( r ) differs in the regions I and II little from -^— Xo» CENTRAL HELD OF FORCE 207 and in region III is in general not of any importance for finding cp(£), since Xo(0 decreases exponentially for r>/v We have thus »W-^fx!<r)*_dffl. A simple transformation leads to h 1 ? 2 (£) where '*«(£-£*)»+»• 4x2 and },2 E — Eq = — k A« *-&&'"»-'>«+"* 8fi % 3 A 3 If we perform the integration in equation (4) we find the decay law _t_ W(t) = e *. The probability to find the particle still in the initial state inside the barrier W(t), decreases by a factor e during a period 15.* We write the SchrSdinger equation for the stationary state of a ft -meson in the field of a nucleus of charge -|-Z* (in this case the motion of the centre of mass of the system is of no interest and can be eliminated in the usual way) in the following form : MM + TprlE — tfCOWW-O. (!) (We note that the use of the non-relativistic wave equation is justified since, as we shall see in a moment, the kinetic energy of the meson is of the order of magnitude 10 MeV, which is small compared with its rest mass, pc*~ 100 MeV.) In equation (1) r is the radius vector of the meson with respect to the centre of the nucleus, j*' = ~tm is the reduced mass of the 208 ANSWERS AND SOLUTIONS system of the meson and the nucleus, E is the energy of the relative motion (since even for light nuclei _3 ucl > 10, j/ is practically the same as the meson mass \i\ we shall, therefore, everywhere in the following omit the prime on the jli). The potential energy U(r), which we shall normalise by putting it equal to zero for r = oo, will be of the form £/(/•) = _Zf_, 3_ 2 R*J r for /•</?, for r>/?. (2) (Expression (2) follows at once, if we take into account the relation U(r) = -#f(r) from the solution of the elementary electrostatic problem of finding the potential <p(r) inside and outside a uniformly charged sphere. ) The potential curve U (/•), and also the energy levels of the meson are give in Fig. 35. We can see from equation (2) that the fi meson moves outside the nucleus in an attractive Coulomb field corres- ponding to a charge -\-Ze and inside the nucleus in the potential of an isotropic Fig. 35 harmonic oscillator of frequency -/ Z£2 HAT 3 * (3) This follows immediately by comparing the oscillator potential Z&* 3 Ze ^ with the expression (2) for the potential, ^ ; the term — -^ in this potential is clearly an inessential constant term. We note that for all real nuclei this characteristic frequency is approximately the same. Indeed it is well-known that v * (4) where \ so that 1.2 • 10~ 13 cm and A is the mass number of the nucleus, 2^ /? 3 z^ A » CENTRAL FIELD OF FORCE 209 but the last ratio is approximately constant for all nuclei (and practi- cally equal to %). In agreement with this we shall assume that the ratio iL is constant when we consider later the limiting cases of Z -► and Z -► oo . The solution of equation (1) for arbitrary values of R is rather cumbersome; we shall not obtain it but restrict our- selves to consider two limiting cases (and at the same time we shall restrict ourselves to the bound states of the meson which are of the greatest interest and from which clearly the discrete spectrum of negative values of the energy E follows), (a) R -> and (b) R -> oo (while j| = constl). The first case (a) corresponds to a "hydrogen-like" system of the nucleus + meson, so that the energy levels of the meson are jiven by the Balmer formula ^* ^~~~mrt (»=1. 2, 3,...). ( 5 ) and the wave functions have "a hydrogen-like" form (with the electron nass m replaced by the meson mass p). In the second case (b) the system is that of an isotropic harmonic oscillator whose energy levels are given by the equations, E* ae = hu>(n + ^) (n = 0, 1, 2, 3....). (6) vhere u> is determined by equation (3). Let us now write down the normalised ground state wave function )f the oscillator ind also the average value of the radius r in this state, oo JrH k |i4«**-4j/i. (7) One can easily verify that for real nuclei (for which R is finite) he region of applicability of the case (a) is determined by the inequal- ty R <d a , where 210 ANSWERS AND SOLUTIONS is the radius of the first Bohr orbit of a meson of mass |* in the field of a nucleus of charge Ze, while the region of applicability of case (b) is determined by the inequality R^> a. The first of these statements is obvious; it means that the dimensions of the "oscillator" region are small compared with the distances which are of importance in the motion in the attractive Coulomb field. The second statement can be obtained as follows: "the oscillatory behaviour "means that the limiting radius R is large compared to the dimensions of the region of motion, — y — ; (this characteristic quantity which is the so-called zero point amplitude of the vibrations of the oscillator is, as can be seen from its meaning, the distance r over which the potential energy %|u** is of the same order of magnitude as the zero point energy of the oscillator, % fio>: V 1>.<D ¥ WO) The quantity r is also, of course, of the same order of magnitude) but from the inequality R ^> y jjs lt f °l lows also b Y using equa- tions (3) and (8) that R^>a. We examine how far each of the limiting cases (a) and (b) are applicable to real nuclei for which we shall evaluate the ratio — . From equations (4) and (8) we have R _ r A^Z^ _ j0 y, p. /■„ „ 4/ /qv where we have introduced for the sake of simplicity the electron mas m and where fl o = ^^°- 53 * 10 " 8cm and where we have also put A « 2Z which is sufficiently accurate for our estimate. Substituting ^^2.3.10 -6 and ^^210, we find that — =1 for Z?«45. CENTRAL FIELD OF FORCE 211 For light nuclei, that is Z< 45 we have thus the "hydrogen- like" case of the motion of the \i meson. The other limiting case z:> 45 (oscillatory behaviour) begins to be of importance and then only to a rough approximation for very heavy nuclei with Z«80— 100. If we put A = 2Z and evaluate r by using equation (7) and (3) we find r^i 7.10" 13 cm: this quantity is not small but approximately the same as the radius of a heavy nucleus from which it is clear that we have still not a limiting case but an intermediate one. Let us note that the solution obtained of the problem gives for nuclei of Z«80 — 90 a characteristic distance between the lower levels of about 6 MeV (this result is independent of any approximation) while the oscillatory Sw is approximately equal to 10 Me V, as can be seen from equation (3). That part of the normalisation integral J I <h> l a dV = 1 (for the ground state) which refers to the region r < /? is approximately equal to 0.55 so that even for heavy nuclei the [i- meson is roughly speaking about 55% of the time inside the nucleus. Finally we see from the level scheme that our assumption about the non-relativistic character of the meson motion was justified. 16.* The probability for transition from a state characterised by the quantum numbers n v l v m x (energy, angular momentum and z-component of angular momentum) to a state n v / 2 , m 2 is propor- tional to the absolute square of the corresponding- matrix element of the V operator, V«ita. njjm, = J <lWi»»i Vtynjt t m t dx, qj where dx = dx dy dz. The matrix elements (1) are those of the gradient operator (or, apart from a multiplying factor the momentum operator) in the E, M\ M z representation. According to the operator equation V = ¥" = ¥{"■ the matrix elemental) is proportional to the matrix element of the velocity operator r but it follows from a well-known matrix equa- tion (see problems 35 and 37, section 3) that £ £ (O^n, = i — l —j- — - (r) nin% , where £„, and £„, are the energy levels, so that the matrix of the gradient reduces to the matrix of the radius vector r, J 4w.ni, VKr im , dx = "' sa w ' ** J 4w»i/4w,»», fo- ( 2 ) 212 ANSWERS AND SOLUTIONS It is thus immediately clear that in the matrix of the gradient (or momentum) only those elements can be different from zero which are non- diagonal in n, that is, in the energy, and the average value of the gradient or of the momentum of a particle in a stationary state of the discrete spectrum (which is the diagonal matrix element in the energy representation) will be equal to zero. (This fact is clearly connected with the fact that the motion in the system is finite: if /> av ^o the particle would go to infinity). In the case of the con- tinuous spectrum, when the motion is infinite, that is, proceeds to non-bounded regions of space, the integral in expression (2) diverges and our conclusions do no longer hold. For a plane wave i i ~e i * r > n the average value of the momentum is given by Pav -P=£<>- If E ni ^E nt the gradient matrix (V) obeys the same selection rules in / and m as the matrix r, namely: the only matrix elements which are different from zero are those for which h— 1 2 = =£ 1. while m l — m 2 = (selection rule for z and Jl ) or m i — m 2 = ±\ d , d . (selection rule for x, -^, y and -^y 6. MOTION OF PARTICLES IN A MAGNETIC FIELD 1. An electron will describe a spiral in a uniform magnetic field if we consider this motion in classical mechanics; the axis of this spiral will be in the direction of the magnetic field. The motion in a plane perpendicular to the magnetic field will be periodic with frequency equal to twice the Larmor frequency (<•> = -|^-) . Let us consider the motion of a wave packet in quantum mechanics. The Schrodinger equation for a particle in a magnetic field can be written as follows:- To find a solution to this problem it is convenient to go over to a rotating system of reference x = x' cos vat' — ysinutf', y = a/ Sin (o£'-|-y COSotf' , 1 ry\ z = z' , t=t', MOTION OF PARTICLES IN A MAGNETIC FIELD 213 W(x, y, z, t) = W(x', /, z', t'). We have then d d i / ^ ' d \ k , The Schrodinger equation in the new variables has the following form: The solution of this equation can be obtained by separating the variables x', /, *'. The equation for the function <p (z') describes free motion along the z-axis. The solution of the equation deter- mining the function ty(x, y, t), is of the form Here x' and y are functions of the coordinates x, y and the time deter- mined by equation (1), the x n are the wave functions of the har- monic oscillator and the A nm axe coefficients which must be chosen in such a way that the initial conditions are satisfied. This expres- sion for <j» (x, y, t) changes sign only if t increases over a period of the classical motion T = -J . Indeed, x' and y' change sign and if we take into account the property of an eigenf unction of the oscillator, *.< — *> = <- 0"X»«>. we get = £ A nm (- i)«x n (*' /^) (- d-x. (y j/f) x In a plane perpendicular to the magnetic field the wave packet will thus change its form periodically with a period equal to the period of the classical motion of a particle in the magnetic field. Along the magnetic field the packet will spread exactly in the same way as for free motion. The wave function <|» (x, y, t) can be found explicitly if we take for the initial, wave function <]>(*, y, 0) = e 2 L 1 h h 214 ANSWERS AND SOLUTIONS If we put all A nm = 0, except A^ = 1 , we find that such a wave packet does not spread in the xy -plane, and that its centre of gravity des- cribes a classical trajectory. 2. To find the operator v it is necessary to take the commuta- tor of the vector r with the Hamiltonian h since h=j-{Hr — rH); we find * /- e A We can now find the commutation rules for these operators e [* 2 c \dx dy J ' (x a c v x v y — v 9 v a = ~ [ — (p x A y — A y p x ) + (p y A x — A x p y )\ = ieh id A y dA x \ leh By cyclic permutation we obtain the other two relations. 3. Let us take the 2-axis along the direction of the magnetic field, the intensity of which we shall denote by &6. The velocity components of the particle satisfy the following permutation rules (see preceding problem), * * A A i@fl A* A A. A* AA v xV v —v v v a ^-^S^, v y v s — v s v y = 0, v g v x — v x v g = 0. The energy operator is equal to « = — + -g-H — §-• We shall put f} in the form of a sum of two commuting operators, "i = "2 ' — 2~ ' 2 — ~2~ ' The eigen values of H axe equal to the sum of the eigen values of H x and H z . Let us first find the eigen values of H v We introduce a new notation v x = aQ, t>„ = <xP, where a = y e -~ . In the variable P, Q the commutation rule is of the form PQ — QP = — t, and the operator // t is given by H x ^=h-^ X (P 2 + Q 2 ). From problem 5, MOTION OF PARTICLES IN A MAGNETIC FIELD 215 section 1 we find the eigen values of H x E in = fi ej ?~(n+i/ 2 ) (« = o, 1, 2,...). The eigen values of H 2 form a continuous spectrum. The energy for the motion in a magnetic field is thus given by E ™° = * jr ("■ + V2) + ~r - 4. We shall take the *-axis along the direction of the magnetic field and the *-axis along the electrical field. We shall take the vector potential of the magnetic field in the form A y = e ^ x , A x = A z — 0- The Hamiltonian operator can in that case be written in the following form : 2(x ^ 2JI - + ljr~ eU ' If we introduce the following notation, ^-X — D — £&■ = « we get for H the expression 2(* ' 2|i. &6 2&e* H" 2,* ' The commutation relation between p x and « is Px* — *Px = — ' — :; — • We find thus that the eigen values of the operator H =^L-u~ are 1 2|i, ' 2p. the same as the energy levels of an oscillator vibrating with twice the Larmor frequency, Since the operators p y and p e , which occur in the last terms of the Hamiltonian operator commute with H lt the operator // 2 =^i_ _ Py°% fxc2g2 2 f* can be put on diagonal form at the same time as H x The energy spectrum of the particle is thus 2 e — h-^t(n -un \jl- Pz p y°^ V-&& 216 ANSWERS AND SOLUTIONS Comparing this result with the result of the preceding problem shows that the electrical field lifts the degeneracy which takes place for the case of a magnetic field only: the energy levels in an electrical field depend on three quantum numbers. 5 * ^n PyPs (x, y, z) = X/7 "Lr eh \ X e&6 e&6*)\' 6. £ nrofc = £/«) 2 -f^(2/i + |m|+l) + fia,m + Ao, (^ + 1 /2). where *=** (« = 0. 1, 2 « = 0. =±=l.=t=2 * = 0, 1, 2,...). where a> = — (twice the Larmor precession frequency). 8. The Schrodinger equation in cylindrical coordinates p, 9, z is of the form ~~2jl lli*"*" dp 2 ' p dp "r P a dcpaj^" 2{xc ^ d<? ^ fyc* v Y Y We write its solution in the form ♦ <">■ * *)- yk *»•"*' '"'•• We introduce the quantities 7 and 8 In the equation which determines the radial function R (p)» MOTION OF PARTICLES IN A MAGNETIC FIELD 217 we introduce a new independent variable £ = fp 2 , so that ^ + ^ + (_| + X-^)/? = 0, (1) where K ~ If - 2~' The required function behaves as £~ £/2 , as £ -> co and is for small £ proportional to £' '" l/2 . We write the solution of the differen- tial equation (1) in the form /? = e-6/2£i« I/a®®. The function te>($) follows from the equation the solution of which is a confluent hyper geometric function, w = />{-(x_J»j + !), 1*1 + 1,5}. In order that the wave function remains finite it is necessary that the quantity X— |m| 2 +1 is equal to a non -negative integer n. The energy levels are thus determined by the expression *=»^-(«+¥+t+t)+-£- 9. In cylindrical coordinates we have J f = 0, 10. The equation for the radial wave function R goes over to the form *+[$*_,:_£(« +~,j"].-.o. where u = >/"p # and where w 2 — -I is changed to m 2 . (This change is 218 ANSWERS AND SOLUTIONS analogous of the change /(/ + !)-*<* + V2) 2 and can be based on the method discussed in the solution of problem 2, section 5). The expression can be considered to be the effective potential energy for a one- dimensional motion. From the quantisation condition fy r .$i : -*!-£(.+Sf) , *~*<«+ , « pi we get the energy spectrum The energy calculated from the minimum of U e{{ (p) is equal to and is the energy of the radial motion, while the energy corresponds to the energy of the rotational motion. The transition to the classical circular orbit is realised provided Ef <C E" or n ^ m 4- 1 '" I . This condition is clearly only satisfied for positive values of m and can thus be written in the form n<^m. 11. If we put the expression under the square root sign equal to zero we get for m > , 12 14. The Pauli equation is of the form MOTION OF PARTICLES IN A MAGNETIC FIELD 219 where H = ^[p—-jA) -\-eU, and where |i is the magnetic moment of the particle. We shall write the wave function in the form The function <p is a solution of the equation For the spin function (*M we then get the equation <4(:;)=-»o(;k)(:;). 15. Since $@ as =$V y = 0, ffl t =$e(t), we have The solutions of these equations are of the form t -^2 (<3V(t)dt S l == C i e t s 2 = c 2 e ° The constants c x and c 2 can be found from the initial conditions, c l = e- ia cosb, c 2 = e ia sinZ. It follows from the form of the functions s t and s 2 that the probability of one or the other orientation of the spin along the z-axis does not change with time. The average value of the x -component of the spin follows from the equation t I a , = -|sin28.cos(^ J &e{t)dt — 2a} and similarly s y = — y sin 28 • sin ;in{^ j 3V(t)dt — 2a}. o The direction along which the spin has a component equal to + £ is characterised by the polar coordinates 8 = 28,$ = 2 (a — ^- I g%> (t)dt\. 220 ANSWERS AND SOLUTIONS This direction will thus describe a conical surface in time. For a constant field strength the line "along which the spin is directed" will rotate uniformly round the direction of the magnetic field with a frequency ^ . 16. The states for arbitrary polarisation of the incoming beam can always be considered to be a super position of two states, one of which, (q \ , has a spin along the -M -axis while the other, i^\ has a spin in the opposite direction. Let us first of all consider the case when the neutron spins in the incoming beam are directed along the 2-axis. The incoming, reflected, and diffracted waves will be of the form ^(i)**- b {\Y'"'' c {\y ,f - The quantities k, k v k 2 are connected with the total energy E and the magnetic moment p of the neutron as follows: From the condition of continuity at the scattering plane (x = 0) of both the wave function and*ttfc derivative with respect to x it follows that: k x A^&&0*==k 2x €. From these relations it follows that k lx = — k x , that is, the angle of incidence cp is equal to the angle of reflection y v For the sake of simplicity we shall put k v = 0. Solving the equations we have (JL\ __ k *~ k ** (£\ = 2k * \A)~~ *» + *** \A) k x + k «2a? — *a> 1/ h 2 kl I+tStM^ The coefficient of reflection R is thus equal to n — ( B \ 2 .( *x-**, \ 2 k ~\a) —Kkv+kn) ' If the neutron spins are directed along the — z-axis we get ^j/l-^grt^'. MOTION OF PARTICLES IN A MAGNETIC FIELD 221 and the final result remains the same. Since jx is negative for a neutron, the angle of diffraction <p 2 * > <p > ?2* (see fig. 36). In the case of an arbitrary orientation of the neutron spin the wave function in the region x > will be of the form C *(J)«"**'+^(?)« < V. where C^. and C+ are the expansion coefficients of the initial spin state in terms of the states ( j and (A. A simple estimate shows that even for $e ~ 10* Oe an appreciable reflection can only take place for very slow (thermal) neutrons (X — 1A) and for angles of incidence cp which differ from j only by a fraction of a degree. Fig. 36 17. The Schrodinger equation for the spin function in the z -representation, (J 1 ), is of the form dt\s 2 ) r Wa> , — /<»; ')(:;) where ^ is the magnetic moment of the particle. We shall introduce the notation ■£-3Vcos§ = a, £<§&dnb = b. In this notation the equations which determine the components s t and s 2 , have the following form: ^ = ibe M s l — tas i . The solution of this set of, equations is where 222 ANSWERS AND SOLUTIONS The quantities A and B are determined from the initial conditions and the normalisation condition | s t | 2 -f- 1 s? ! 2 = 1 • After some simple calculations we get for the transition probability the following value : *(7 ■ -4) ~ ,+«■-£.,» «"* [I » (' - 2, cos » + «"■] . where q is the ratio of the Larmor precession frequency to the frequency w of the rotating magnetic field, 2[A(ffff _ wo The quantity ? is positive if the magnetic field rotates in the directio of the precession and negative if the rotation is in the opposite direc- tion. If the angle is small, that is, |- ^<1, the transition probability is approximately equal to From this formula it follows that at resonance a> = u) , or for q = -\-\ the probability for a re -orientation of the magnetic moment with respect to the magnetic field, which is equal to P (-^ , — -j)^ « sir. 2 y », can become practically equal to unity for certain values of t. If in the case under consideration we change the direction of rotation of the magnetic field, or change the sign of $B Z we find for the transition probability the result f(l.-l) = T i! "' which is considerably less than unity. On the basis of such a sharp qualitative difference one can determine the sign of the magnetic moment of the particle. 18. a) MOTION OF PARTICLES IN A MAGNETIC FIELD 223 b) 2(0 = (z) + ^2(aa*-pp*), csg-(53+ +3J&)'. rf-c. Note. We can consider, for instance, a system of particles the 2-components of the spins of which at / = have definitely the value +^, that is a— 1, p = 0. From the results we have obtained, one easily sees that if there is screening along the track of such particles when they are moving in an inhomogeneous magnetic field, these particles will form two spots. The 2-components of these spots will be the same but their ^-coordinates will have opposite signs. 19. The direction of the magnetic field will be characterised by polar angles ft, cp. These angles are functions of the time. The Hamiltonian operator for a neutral particle can be written in the form H = — jx &6 ij 9 sin cos <p + J y sin sin <p + K cos &)"» where &6 is the absolute value of the magnetic field intensity. We shall denote by \ the angular momentum operator in the direction of the magnetic field J s = J x sin ft cos 9 -f- J y sin ft sin <p -j- J z cos ft and we shall introduce a function <W»(0> which is the eigenfunction of the operator y £ , or, We shall write the solution of the Schr&dinger equation in the form 224 ANSWERS AND SOLUTIONS It is well-known (see problem 19 section 4) that ^ m (t) = e' iJ ^e- iJ V^, where »J<°> satisfies the relation We shall first of all evaluate ty m (t). If we use the relations e iS v*J z e-^v* = J z cos — J x sin 0, W = y ^ + »)C/-*+l) *£?-i + — Y^y+m+ixy— »)C+i. we have $m (0 = ( — *? w C0S ft ) *» (0 + + 1 V(7+«+i)(y— ») ('? sin ° — ^) Wi (0 + -f • -g- V r (;+»)(/-»+ 1 ) ('? sin ° + ;c> ) **-* W- If we substitute this value of i? m (t) into the expression to S (<K» (0 «» (0 + «» (0 L (0} = — f^ 2 »«» (0 <W» (0. m m we obtain a set of equations which determine the change in the coefficients a m with time, ih ^ + m^&€a m = — mfi<j> cos §a m + + i'fi (<f sin + id) V r C/ + »)0 , -»+l)'»»-i + + 1 A (y sin » — /») VC/ — w)C/H-w+l)«m+r If that is, if the angular velocity of the change of direction of the magnetic field is much smaller than the frequency of precession, we can neglect in this set of equations the right hand side and find thus ATOMS 225 In this case the probability for different values of the angular mom- entum components does not change with time. 20.* If we write the Hamiltonian in the form "-J^-HX*— ?4 we verify easily that the condition that it is the same as H' is that the same components of the vector operators p~-jV and A(x, v, z) commute, that is V4 == div 4 = 0. 7. ATOMS 1. From the given inequality we get J|V<>| 2 dt + Z J(VK| 2 Vr)rfx+Z2j*(Vr) 2 K| 2 ^>0. Integrating the second term by parts and using the fact that (V r) 2 = 1 , o and Ar = --, we have r The left hand side of the inequality is the average value of the Hamil- tonian operator (in units e = h = v- = \ ) H=— h-A — — for the state ty. The lower limit of the energy, — ^ , is reached for the state with wave function %, which satisfies the first-order differential equa- tion, V*o-fZ<i> o Vr = 0, from which it follows that <|»o~«-*. 5. We evaluate first of all the wave function in the momentum representation from the general formula For a Is -state we find ?i. , x 1 /2a\'/» 1 226 ANSWERS AND SOLUTIONS Similarly we find for a 2 s -state . . 1 (2a\Vt fi a 4 For a 2 p -state there are three eigenf unctions (m g == — 1, 0, + 1) ^p W y2«\fii t //> 2 a 2 . 1\8 ' HtF + TJ From these expressions we can find the normalised momentum distribution w (P) = |«P (/>) I 2 - 6 . v r ^i — ? _ Vfl»(/i a + 2W </ + !)' r r r — 2 arises For a given value of n the minimum value of this expression es for "circular orbits", that is for l = n — 1 2 r Y2n + 1 7. If »i=l, » 2 = 0, w = + ^ 2= « 21 (r)r 10 (», cp). 9. The Hamiltonian function in relativistic mechanics is of tin form where h —- ^~ We shall now take p to be the operator p = — ihV, and // will be considered to be a small purbation. We then get in the approxima- tion used by us in the Schrodinger equation \p + U (r) i ty = E<V» and the ATOMS 227 required correction for the energy in a state with n, /, m is 2^2 «Vc2(2/+l) |_8/i4 (2t+l)n*}'W\$i) • 10. Instead of starting from the unperturbed wave functions with well-determined values of l s and s z and later on solving the secular equation it is convenient to choose as initial eigenf unctions with well defined values of P and p, where /=/-f £ is the total angular momentum which can easily be shown to commute with H % . If we take into account that for these functions the following relation holds P*=JU+ !) = /(/ -h !) + *(«+ l) + 2to f we find that A£ 2 = 77 2 -^Q'+l)-/(/ +l)-s(s+l) ft2 7T5IA 2 2 VP * V7dF/- For the hydrogen atom u = — — and since " „» ( , +1) (, + ^>i' we find finally This formula can be written more briefly since in our case 5 = £ and we have either j = l — % or y = / + V 2 - One can easily show that 2fe=./CH-l)— J(H-1)— s(H-l)= so that for all values of j and / ' if J=l+\ — (/+i) if y=/— j, V J -hi- / + i 2 ^2 228 ANSWERS AND SOLUTIONS Combining AE 2 with the correction which describes the dependence of the mass on the velocity, AE t , (see preceding problem), we find \F — ^t( e l\L(l. L-\ c — tfl \hc) ri* \8n 2/ + 1/ ' This expression does not depend on /, which means that two states with the same value of j but different / correspond to the same (degenerate) energy. 12. The tritium nucleus goes over into the hie nucleus during p -decay. The influence of 0-decay on the atomic electron is essentially contained in the fact that during the short time interval, t <C ~i » the potential energy of the electron in the atom changes becomes equal to U = — — instead of U = -. The time t can be estimated as the time of flight of the 0-particle through the atom, a t v fi* where a n = —„ v is the velocity of the R-particle. Since the (AC' energy of the p-particle is of the order of several keV, we find t oi -^— The wave function of the electron does not have time to change during t, since we get from the Schrddinger equation Let us expand the wave function of the electron ty in terms of the eigenf unctions of the electron in the field with Z = 2 , ra The expansion coefficients, Ck = j <]">* dx determine the probability for excitation, "n = 2l Cnl ATOMS 229 and ionisation ™ion = J \c b \ 2 dk. Since «j» is spherically symmetric, c n and c k will be different from zero only if the state n, k is an s- state (/ = 0). Since we find If we put Z = 2, Z' = l, we get for n = 1: 16 yT c, = 27 ' that is, the probability for the ^e-ion to be in the ground state is equal to w x = \ q | 2 = (-|) 3 = 0.70. The total probability of excitation and ionisation will thus be equal to 1 — w x = 0.30. For n = 2, c 2 =— 2"' W 2 = 0.25. Using the equation /="(«, p, T , x) = (l — jf)T— Pf ( T _ a, T — p, T> AT) we find — I 1 2 _ 29 " 5 (n — 2)2"-* We find for the value of the probability for excitation of the first few levels, using this formula, Wl = (8\\ w 1 w = 2»35_ 1 \9/ 2 4' w 3— 5W — l ' 6 /o> ™4 = 2^-0.39%. 13. The Hamiltonian is of the following form(we shall use atomic units e = h = \>.= l) // = _±A — 1a — z — Z -4- ! 2 * J 2 7^ 7"~i~7~- ^ z 'l r 2 '12 230 ANSWERS AND SOLUTIONS According to the variational principle it is necessary to evaluate the integral E (7J) = J ¥ (r v r 2 ) H\ (r v r 2 )dx x dx 2 dE and to determine the quantity Z' from the condition _- = 0. In our case we have where c — - . The integrals of the first four terms can easily be evaluated, /^(r 1 ,r 2 ){-lA 1 -iA 2 -f-|}x XWv r 2 )dx l dx 2 = Z' 2 -2Zr. As to the integral involving J- , it is convenient to evaluate it in elliptical coordinates, s = r 1 -\-r 2 , t = r x — r 2 , u = r l2 , dx l dx z = it 2 (s 2 — t 2 ) u ds dt du, — «<*<«, 0<«<s<oo. The result of this calculation is f^ 2 (r v r 2 )~dx 1 dx 2 = = u 2 c 2 f ds j du J dte 6 o -u We get finally £:(Z') = Z' 2 — 2ZZ' + |z'. From minimising E(Z') we find z —*< 16 For this value of Z' we get for the ground state energy *—(*-£)"• To find out how accurate our result is we calculate the ionisation potential of helium ( Z =2) and compare the result with the experi- mental value. The ionisation potential of helium / He . is equal to CO 8 +« ATOMS 231 the difference of energy of the once ionised helium atom and the neutral helium atom in its ground state. The value of / He is equal to (1 at. un. = 2Ry = 27 eV). / He = 0.8476 at.un. = 1. 695 fly . The experimental value is 1.810 Ry. The ionisation potentials of other systems with two electrons, Li , Be , and so on, are also known experimentally. Comparing them with the theoretical value we get Element II N X II N + II N + + o> 03 II N + + + 03 w + + + + '<*»,„• ■ ■ 1.6952 5.445 11.195 18.945 28.695 7 <«»«p 1.810 5.560 11.307 19.061 28.816 The results of the evaluation of the ground state energy is in satisfactory agreement with the experimental values. 15. Both the orbital and the spin angular momentum are equal to zero in the ground state of the helium atom. As a consequence helium is diamagnetic. The diamagnetic susceptibility evaluated for one mole is given by the expression where r?+^ = J"(r?-|-r 2 2 ).f rfT lfi ?x 2 and where N A is the Avogadro number. The approximate expression for the ground state wave function of the helium atom is *< r »' r 2)=ir3 g - z ' (n+r ' )/a - If we evaluate the average value r\-\-r\ using that wave function we get: 1' 2 7/2" 232 ANSWERS AND SOLUTIONS Substituting this value into the expression for the diamagnetic sus- ceptibility we find x = — -1.67- 10"'. The experimental value for the diamagnetic susceptibility is given by x = — (1.90 zt 0.02)- 10" 6 . 16. It is convenient to introduce the following notation: Z 1= =a, Z 2 =2p, 2Z; /s = a, cZ*t = b. From the fact that the functions are orthonormal we get In the new variables <K and ty 2 have the following form : fc = <|> 100 = ae-^Yoo, ^ 2 = ^oo = * [l — 1(« + P) rjf-»r w . The approximate ground state wave function of the lithium atom can be written as follows: <W ( 1 ) % (°l) <W ( 2 ) % (°2) <W ( 3 > "*>+ ( °3) ^(l)'n- (0i) <W(2)fl- (a 2 ) ^(3)ti_ (a 8 ) <fe ( 1 ) % (°i) <tt ( 2 ) % (°*> ^ < 3) *"+ ( ° 3) where %(^)= 1, %(-y) = 0. ^-(tH ' ,| >-(— f)^^ In this state S = V 2 > M = V a - The Hamiltonian operator in atomic units ( e = h = i>. = l) is for our problem of the form /3! i = l We shall evaluate the energy in the state $. The kinetic energy of an electron in the Is -state is equal to and in the 2s -state T.-5 + a 2_ap + £2 ' ATOMS 233 The interaction energy of the inner electrons with the nucleus is U 1 = ~ I -ydx = Za 2 J e — F -r*dr = — Za, and of an outer electron with the nucleus (for lithium Z = 3) Zp Zf a-2p U 2 01 2 r 2 aa-ap + p2 The energy of the Coulomb interaction of the inner electrons is equal to Kn = J* J 4 ^ (r ° |2 • +» (r2) |2 dti dt2 = t a - The interaction energy of the outer electron with the inner electrons is equal to 2AT 12 = 2 J j* i- 1 ^ (ri ) |2 1 ^ (r2> p rfTi dT2 = The exchange interaction energy of the two electrons with parallel spin is oo A = 2a 2 * 2 j" <r (a+f}) r « [l — j (a -f- p) r 2 1 r\ dr 2 X oo _ 4a3p5 If we put p = Xa, we get 27 1 +r 2 = r = a 2 c Pl (X), 2t/i -h U 2 + AT n + 2/C 12 — i4 = — acp 2 (X), £ = a 2 T 1 (X) — acp 2 (X). The minimum value of the energy is realised for values of a and X which satisfy the equations ^ = ^ = or da <?X 20^ (X) — cp 2 (X) = 0, acpi (X) — cp 2 (X) = 0. 234 ANSWERS AND SOLUTIONS Eliminating a, we get ! ^)_2 ! ^) = 0) X=0 .2846. The corresponding values of a and p are a = 2.694, p = 0.767. If we substitute these values of the variational parameters into £, we find for the ground state energy of the lithium atom £ = — 7.414 at.un. or, £ = — 200.8 eV. Experimentally £ exp = - 202.54 eV. Applying perturbation theory, that is putting a = 3, p =•/„ we get for the ground state energy a less accurate value than from the variational method, namely £ = -7.05 or in eV £ = —190.84 eV. 17. Let us consider the operator of the nuclear kinetic energy, f. In the centre of mass system where P is the nuclear momentum and p i axe the electronic momenta, 7" is of the form i > k Since the ratio of the electronic to the nuclear mass J<1, we can write the required energy shift using perturbation theory as follows : A£ = j 6*7^ dx, where y is the wave function of the electrons in the field of an infinitely heavy, fixed nucleus. The first term in the expression for f differs only by a factor JUL from the kinetic energy of the electrons which is equal to the M energy of the atom with the opposite sign by using the virial theorem . We get thus for A£ a sum of two terms: A£ = A£ 1 4-A£ 2 , ATOMS 235 We shall now consider in more detail the term AE 2 . If we take for <V a product of wave functions of the separate electrons, AE 2 will tend to zero, since the average value of the momentum of an electron in a bound state is always equal to zero. If, however, we symmet- rise such a wave function in the appropriate manner, A£ 2 will be different from zero. The symmetrised wave function of the helium atom can be written in the form * (ri * rz) = 71 l<fi (ri) * 2 (rs) — ^ (rz) *» (ri)1 ' where the upper sign refers to para -helium (total spin S= 0) and the lower sign to ortho -helium (5 = 1). If we put this expression into the equation for A£ 2 , we get The matrix element of the momentum is different from zero only if A/ = 0, ± 1 , and AE 2 is therefore equal to zero for \snd, Isnf , . . . , states. If we take for the ls-electron and for the np-electron hydrogen wave functions with effective charges Z l , Z 2 ^s(r)=y^e-z>r t 2ZlVn i Z~r ^ P =r i0 -3- 2 n2 re »/>(_«+ 2,4;^). we get where the upper sign refers to paraterms \snp l P, and the lower sign to orthoterms lsnp s P. 18. Let <l»(jf, v, z) be a solution of the Schrodinger equation referring to the discrete energy spectrum. Let us consider the one parameter set of normalised functions of the form x' / ^(Xjc, Xy, Xz). The expression -\-U(x, y, z)\ty(kc, Xy, \z)f\dxdydz, considered as a function of X, must have an extremum for X = 1 , that is 236 ANSWERS AND SOLUTIONS If we introduce new variables of integration }.x, t.y, }jz, we get /(}.)= *. 2 f-|- >.-*{?. Hence we find 2T— v(7=-0. The virial theorem can easily be generalised to the case of a system of many particles. 19. a)Z- v '-^. b)Z'"£.* c)2 4/ ^. d)Z'^ ( e) Z*£ . f) Z''"*. g) z''\ 20. The total energy consists of three parts, the kinetic energy of the electrons T, the interaction energy of the electrons and the nucleus U^, and the energy of the interaction of the electrons with one another, U ee . The last two terms have the following form, U _lf Pir)P(r') dxdx , U ™— 2 J \r-r'\ axax • To evaluate the kinetic energy let us consider an infinitesimal volumi element of the atom, dx. The number of electrons with momenta between p and p-\-dp, is proportional to the phase volume and equal to: 8rc/?2 dp di p 2 dp di The electron density is obtained by integrating over p from to some maximum value p = P > ft The electronic kinetic energy in the volume d: is equal to : rfi dT= I ?dn = j^dx. If we express in this equation p in terms of p and integrate over the atomic volume we find the kinetic energy of the electrons r =w^ 3ir2 ) ,/3 jV /;,rfx - ATOMS 237 We find finally for the total energy E^T + Un+U^ 22. The volume element expressed in terms of x is of the form d- = 4-r- dr = 8-/Av 5 </.v. We can evaluate the kinetic energy and the energy of interaction between the electrons and the nucleus, U m =*—SkAWZ. To evaluate U ee = ~ J J p ( ( r r) _ p ^f | } rfx dx', we find first of all the poten- tial produced by the electrons, cp e . By solving the Poisson equation Acp e = 4irp, we get ?. = ]j-N -«-*(*-+- 1)1- If we now use Green's theorem to evaluate £/„« we find : = — "2 J ?«P rfT = * The normalisation condition determines 4 , J p rfx = 16tc/4X 3 = A/. Substituting this value for A in the expressions for T, U^, U M , we get T _ 12 /37cAT\% l 25tcV 16 / Xa ' . u «« 2A. * 238 ANSWERS AND SOLUTIONS The minimum of E = T+U W + U W is reached for . _ 9 (Z*N \v. l *• — 2b\W) -_N_ * 8 and is equal to For a neutral atom we get F- 2 ±^- (^V /3 Z' /3 ==0.758Z % at.un. c — 36 • 64 \3n) 23. Let p(r) be the expression for the electronic density in the Thomas -Fermi model. In that case P (r) will lead to a minimum of the energy of the atom, If we substitute in that expression instead of p the function X»p(Xr), which satisfies the same normalisation condition as p (r), we find E( l )== \2T + W, where T is the kinetic and U the potential energy of the electrons in the atom. Since E (X) must have a minimum for X= 1 we find that the following equation must hold, 2T-\-U = 0, which is the virial theorem. 24. The energy of the interaction of the electrons can be written in the form U ee = -jj<?e?dr = ~jt dx -^ 9P ch, (1) where cp e is the potential produced by the electrons, and cp is the potential of the self -consistent field, including the field of the nucleu In the Thomas -Fermi model the following relations are satisfied, A o Pi ATOMS 239 where /> is the limiting value of the momentum, and <p the potential at the boundary of the atom. If we eliminate p and express <p in terms of p.we find i/.-4j**-2£>*-¥ The first two terms differ only in their factors from the energy of the interaction of the electrons with the nucleus, and the kinetic energy, We have thus Uee = ~jU nB -±T-*» N 2 " If we now take the value of T from the virial theorem, we find finally For a neutral atom (N = Z) <p = and 25. The energy of total ionisation is equal to the total energy of the electrons, but with opposite sign. Using the virial theorem we find (see preceding problem) We shall transform the expression, U ne = -Z Jf dx as follows: we introduce the potential <p e , which is produced by the electrons A<p e = 4«p and use Green's theorem 240 ANSWERS AND SOLUTIONS (the surface integral over the boundary of the atom is equal to zero and A — = — 4™(r)). r We have thus ^ion=-T Z ^(0) + |^o- If we go over to Thomas -Fermi units, xbZ 1 /3rc\ V3 Z'^W ■ 6== 2 (t) • V-'o— r Jf we find Te(O = ?-| = ?0"X ll -^ W1 ' and since we have for small value of x with a = a = 1 .58 for a neutral atom <for a positive ion a > a ), we have finally _3_£^ 3 3 (Z — N)*Z i/a £ ion— 7 b a 7 bx where * is the radius of the (Z — N) times ionised atom. 26. The potential of a point Coulomb centre is the same as the potential of a sphere charged on the surface outside that sphere, if the total charge is the same in both cases. Inside this sphere the difference between the two potentials is given by The change in the potential energy of the atomic electrons is equal to «/ — z^fj(i-i).W. i = l where we have introduced the auxiliary function (1 r<a, e(r) = }0 r>a. According to first order perturbation theory the energy shift is N ATOMS 241 If we integrate over all variables except one we get J H(r l ...r y )|*dT 2 ...dx x = s i r p(r), where p(r) is the electron density. In this way we get AE = - Ze* J p (r) (1 — 1) t (r) rfx. We can use the fact that p(r) changes little in the region r < a. If we take this quantity from under the integral sign and put it equal to its value at the origin we get Af = — Z« 2 p(0)^-« 2 . 27. The wave function of an s-electron is where x„ satisfies the equation oo and the normalising condition fy^dr=l. o In the semi -classical approximation the solution of the equation for Xn is of the form \jjp« d r + 9j Xn = y^cos ( -i- | />„rfr + <p ) , (1) where p n =V^[E n —U(r)\. if r is small, (r<C — rr~ — ^) then first of all we can neglect the This solution is, however, useless for small values of r. Indeed, screening of the field of the nucleus and put U(r)— — ; secondly we can also neglect E„ compared to U (r). If we put p = y ** e into the condition of applicability of the semi -classical approximation «( A ) 242 ANSWERS AND SOLUTIONS we get To obtain an expression for Xn which can be used for small values of r, we start again from the original equation and substitute for u (r), If? and also neglect E n : 7 "+%*? 7f|S= 0. This equation has the following solution ; •/.» = c„i/7j l ( 2 /*p). (2) To find the connection between the constants C n and A n , we note that the region of applicability of the semi -classical solution (1), and the region of applicability of solution (2), where the screening of the nuclear field has been neglected, overlap for large values of Z so that solutions (1) and (2) must co- incide in the region <^ r <C! — r, . We shall show that solutions (1) and (2) are the same in the gen- eral region of applicability. To do that we put in solution (1) p = ^l/^f!. We then get A n V"r /2V2iIzPF \ fta m /Q . -/.»=* — - — cos — b? — OO^— • ( 3 ) The condition r^> — — means that the argument of the Bessel function in expression (2) is large. But for- large values of the argu- ment (x^>l) we have The solution (2) is thus of the form n -./- / h (l V 2\i.Ze*r 3n\ Z\t-e l Z '■' ve* ATOMS 243 Comparing this expression with expression (3) we find We must now find ^(0) . For x <; l we have y, ( x ) . 4 and using equation (2) we find 2 Hence, The constant A n can be determined from the normalisation condition 1 I K '» / yldr = A 2 " 7 lf , "J V*l^-U(r)l <tr - _4 f dr '"' 2 •' V^'[£»-£/(r)J =L If we differentiate the quantum condition //2ji[£ B — £/(r)] rfr = tu (n -+- T ) A, which determines E n as a function of n, with respect to n we find ' dn J icA. V 2(i.[£ n -f/(r)J Comparing this last expression with the normalisation condition we get A i =,^: d Mn n ith dn and finally For an unscreened Coulomb field we have and we would then get from equation (4) 42 (0 ) = i- (-££?) s which is the same as the result obtained from a rigorous calculation. In atomic spectroscopy the following equation is often used for the 244 ANSWERS AND SOLUTIONS energy levels of the excited states of the valence electron „ v£ __J ^n— 2\P (/? — =)• ' where a is the so-called quantum defect which depends only weakly on n. If we use this formula for E n we get •> 2 (0) Z (*%« («-£) n A« (n — <*)* 28. Let the electron be in a stationary state with well-defined value m of the z -component of the angular momentum. The wave function of such a state is equal to: ■**.('. ». *) = Rm(r)P ( i m) (cos »)«'"». The electrical current density in the state ty n i m is in a system of polar coordinates, r, 0, <p , of the following form : J r = y» = 0, J.= "-^-l^-l". '* jxr sin § It is obvious that the magnetic field vector will be along the z -axis. A circular current of strength dJ will produce at the point O (see fig. 37) a magnetic field, the strength of which is given by rfje> 2 =^2irsin 2 &; since dJ = J^r Fig. 37 >dr, so that ^? z =- 2it]e]hm ?c oo « °° This result can also be obtained in a different way. It is well-known in electrodynamics that the magnetic field strength produced by the motion of a charge is equal to (we neglect here retardation effects): where r is the radius vector from the point of observation, and / the angular momentum. ATOMS 245 W \_e_]h so that In quantum theory, in order to obtain the energy value of the magnetic field strength, we must evaluate an integral of the form J <!»* -fi ^ dx > since 4<|»=m<|», The average values of S@ x and &6 \ will in our case be equal to zero since For a 2/7 -state (m= 1) we have 29. The magnetic moment of the particle is equal to: In the case of two particles we introduce new variables: the centre of mass coordinates {X, Y, Z) , and the relative coordinates (x, v, z). The average values of the magnetic moment expressed in the new variables will be 2K e 2c and similar expressions for SW^ and 9JFj,. In a stationary state, the average values of the coordinates x, y, AAA z and of the momenta —ihj- x , — ih-^ , — *fi jjj are equal to zero. The above expression simplifies thus and becomes m. =-£('-*) J *•{-«(*£-'£)}♦*• In the present problem m is the mass of an electron and M the mass of the nucleus. Z« 30. A£ = 00844 -2.79 =5- cm -1 . For the hydrogen atom ground state (Z = n=l) we have A£ = 0.0235 cm" 1 . 246 ANSWERS AND SOLUTIONS 31. To determine the energy it is necessary to find the magnetic field strength produced by the electron. Owing to the orbital motion of the electron there will be at the position of the nucleus a magnetic field the strength of which is given by the Biot-Savart law, §Ci = 1 [rj] c where r is the radius vector from the nucleus to the electron and j = — ev (-e is the electronic charge). We introduce the orbital angular momentum operator I. We then get for ICi * eh 1 : Since the electron possesses apart from an electric charge also a spin, the total magnetic field strength at the position of the nucleus will be equal to The operator of the hyperfine structure energy can thus be written in the following form w=— poe, where I is the operator of the nuclear spin, and p/* is the magnetic moment. We shall consider w to be a small perturbation. The un- perturbed state is characterised by the quantum numbers n, j(j = l-\- 1 i j=l— i f 2 ), I (we shall assume LS coupling). To determine the hyperfine structure energy we must average the operator w over the state with quantum numbers If we use equation /, h /</=/ + *). • fm' V/A) . f , we get r an wW, "i - - - (IP /(/+1) - d) Jm i . W) alim . - ^ \ r n ) nJjm ^ j (J + 1) yJhmj Using this relation we can easily show that the operator w can be written in the form: ATOMS 247 It follows thus that the hyperfine structure energy E is determined by the following expression : We see thus that any term characterised by the numbers n, I, j, is split into 2/-+-1 components (if y>0 because of the hyperfine structure. The different values of / give us the interval rule for the hyperfine structure multiplets. 3 z Si I • v.* 0.0025 3*P, ■ v,* 0.00 OS c ■v,+ 0.0583 ■v. 3% D ? - 5B90& em-' • v.+O.OOSJ Y,* 0.0583 V, D f - smf\ Fig. 38 We note that by simple counting the number of components of the hyperfine structure multiplet in the spectrum of a given isotope we can determine the nuclear spin. In figure 38 we have given the hyper- fine structure of the sodium D lines. The fine structure, that is, the presence of a doublet (the line D x = 5896 A, corresponding to a transition 3 2 P»/, -*• 3 2 Si/, and D 2 = 5890 A, corresponding to a transition 3 2 P V , -► 3 2 5i /t ), is due to the spin -orbit interaction (see problem 10, section 7). 32. Not only the total angular momentum but also separately both the total orbital angular momentum and the total spin angular momen- tum of the electrons are equal to zero in diamagnetic atoms. Because of its precession the electron acquires an extra velocity, If we denote by A the vector potential of the external magnetic field A = 4- 13C*]- we can write this last equation in the form v' = — A. 248 ANSWERS AND SOLUTIONS The current density produced by the precession of the electron shells is equal to: •j.c ' where p(r) is the charge density at r (the electronic charge is equal to - e). We first of all evaluate the vector potential A' pro-. duced by the magnetic field, A' l o = &JM T ££ T «. If we use the equation l l r la YR* + r* — 2Rr {cos cos » -f- sin sin » cos ($ — <p) ri A4V 2/ + 1 TT Y] m (0, *) K Iw (ft, cp) # — > !•</?, — . « + l K w (0) /?* I, m 1 fJ+1 . K < r, we get the following expression for 4': Using this expression one can easily evaluate the magnetic strength along the z -axis. This field strength is r' <r r r >r The field at the centre of the atom, that is the field acting on the nucleus, *-.<<>>-£ J ^"-giw- does depend on the electrostatic potential <p (0), produced by the electrons. In the Thorn as -Fermi model we have 9(0) = — 1.588^, where 6 = 0.858 -£-, a =-21 and thus J^ z (0) = — 0.319- IQ- 1 Z''3&. 33. ^(0) = —g^ = _ 0.599- lO" 4 ^. ATOMS 249 34. iS . *S X . »P 0ilil , 2 D Vt , v ., «D, A . ,,...,„,*. 35. a) i5 8 5 i; c) iD 2 3D 123 ; d) iSo'S^P,. *P 012 l D 2 *D 123 . 36. a) *S 2 P*D; b) A S 8 PiD»fiG; C) 2 5 2 P*P2£) # 37. 0*P 2 , CIV,-. Fe'D v CoV,, A.«&,., La'D.,,. 38. K, Zn, C, O are even; B, N, CI are odd. 39. If all triples of quantum numbers are different then the number of states is equal to the number of ways in which we can take W from K+m h , or, S(M S ) ■ iV-i AT. If there are N' pairs of identical triples then we have 40. The number of states is equal to - "'W-O-M-'+D . „ h e re A,, = 2<2, + 1 ). 42. We shall denote the antisymmetric wave function $ = N\ W^f (W * H * t N m X„H (5.) • • • * n K t N m K m S G N ) which is constructed from functions of the one-electron problem, simply by &(nWm\m\, n 2 l 2 m 2 m 2 , .... n N l^m^m N ). Let us consider the action of the symmetrical operator, AT (4— il 9 )= ^(ii—tii) i = l " on the antisymmetric function <&(nH l m\m l . n 2 l 2 m 2 m 2 , .... n N l K m?m K ). 250 ANSWERS AND SOLUTIONS One easily sees that (L x — iL y )$(n l l x m\m\, nYmjml ■•■) = = f ( ;i + w {)(/'_ m ; + l) $ (nH l m\ — \m 9 , nH*m\tn\ t . . .) + + yqt _|_ m p (/ 2_ m * + l ) $ (ni/imj wj, /i 2 / 2 m 2 - 1 m 2 , ...)+••• The action of the operator S x — lS y gives a similar result with, however, m 8 instead of tn t decreased by unity. If the wave function is an eigenf unction of the four commuting operators S 2 , L 2 , S g , L s , then the action of the operator s (L x — i£ y ) and (S x — iS y ) leads to the following result: (4 — iL y ) $ (SLM S M L ) = = V(L-|-Mi)(L — M£+l) * (SZ.M S M L — 1). (S a — /5„)$(SLM s Alx) = (1) = V< > S4-Ms)( > S — M S +1)$(5LM S — IMl)- After these preliminaries we can immediately go over to the solutior of our problem. In the case under consideration we are dealing with equivalent electrons so that we can everywhere omit the quantum numbers nl. The value of the z -component of the spin angular momentum will be indicated by indices (±r) on m x . For the configuration p* we get the following classification of states according to the values of Ms and M L (we limit ourselves to non -negative values) and we get the following term scheme: Ms Ml -L 1 $(1 + , — 1 + $(1 + , — 1 + $(1~, — 1 + $(1 + , — 1" $(1\ 1" $(1 + , + $(1 + , 1" . + ) [ 4 5] . 0") r*s" . o + ) 2 P , 0+) JD. . -1 + ) \ 2p ] , o-) M . + ) [*D] *1 * 2 $3 Afterwards we evaluate the action of the operators (L x — lL y ), (S x — it on each of the states given a moment ago, (L x — iL y ) $5 = V% (*, — *a). ATOMS 251 (L x — lL y ) <J> 6 = ]/ 2 (*,-*,), (S x — tS y ) *, = <f> 2 + <I> 3 -f <I> 4 . where $ t is the wave function of the state *S with Afs = -j , Mj, — 0. The action of the operator (S x — iS v ) leads, according to equa- tion (1), to the result (S x -iS y ) *(«S. 1 . 0)= ^3* (<S, 4- . 0) . (5* — tSy) ^ = <I> 2 + *, 4" ^ Since we have $ (*S, j . 0) = -^=- (4> 2 4" *a + *i). Similarly we find for 2 D terms the following wave function: <fr(*D, 1, 0) = _L-(<J> 3 _2* 2 4- **)• The state Z P, -^ , 1 (Afs == 9- » M L = I ] is a linear combination of the state <£ B and 6 , which is orthogonal to the state 2 £>, 4". 1. From this we get and also *( 2/> ' 2- 'H^f (*5 + *e)> *( 2P ' Y'°) = -^-^3-^)- In the same way we can obtain the wave functions corresponding to negative values of the z -components of the angular momentum. 43. From the table which is given below it follows that to deter- mine the eigenf unctions of the two 2 D terms we must first evaluate those of the 2 H, 2 G, *F, 2 F terms M S M L 1 5 $(2 + , 2", 1 + ) $ t [ 2 //l 252 ANSWERS AND SOLUTIONS M S M L ± 4 ~ 3 ■k 2 'I>(2' f 2" . 0") $(2", 1 + O < I»(2 ! , 2" . -i + ) *(2", 1 + o") #(2 + , 1" + ) ^<2 _ , 1 + + ) *(2 + , 2" -2 + ) *(2 + , 1 + — 1") *(2 + , r — 1 + ) $(2", i + , -1 + ) $(2 + , o H o~) *(1 + , r, o + ) $(2 + , i + . + ) $(2 + , i + , — 1 + ) *2 *8 ** *, - E] -2/y- 2/7 *8 *!« *12 13 — ^ — ■2 H - 2 G 4/7 2/7 2 D 2 D *1 We first of all determine the result of the action of the operators (L x — iL y ), {S x — iS y ) on each of the following states: (Z x — IL V ) * t = - 2$ 3 -f- V 6*2. (£» — 14)*, = — 2* I +2* e + V r 6® 4 . (4 — /Z,) $ 3 = - V 6*6 + / 6$ B . (5„-/5 I ,)* M :=*, + *,-+-*5. (4 — /4) $ 4 = — 2* u + 2$ t0 + 2* 6 . (4— '4)*6=V r 6*11+^6*., (4 — /S,)* 16 = * n + *io + * 9 , (4 — *4)*, — 2* lt +y r 6* 11 . The wave function of the state 2 H, ^ , 5 (m s = ^ , M L = 5) is equal tc 3V or $ ( 2 #, ^ . 5^ = * t . If we apply the operator (4 — *4)> we get: {L x ~iL y )$(*H, I, 5) = 1/10* ( 2 //, I. 4). Since (4 — *4> $1 = — 2<J> 3 -f- /6$ 2 , ATOMS 253 The other states of the 2 H term which are necessary for the solution of our problem can be found similarly: *( 2// « T' 3)=-^=-{/6<!» < -2<I» 6 -h4* 6 -2$ 7 }, *(*"> 2' 2)^ y^{*8-*9+3* 10 -2* ll -3* 1 H-/6* 1 ,}. The state $ ( 2 G, y . 4 j is a linear combination of the states $ 2 and $ 3 , which is orthogonal to the state $ ( 2 H, y , 4j. From these conditions we can determine the wave function $ ( 2 0, -* , 4], Since there are no phase relations between states of different terms we can put a = 0. The other states of the 2 G term can directly be obtained by con- secutive applications of the operator (L a — il y ). We have in that way *( 2 G, j, 3) = -^{/6* 4 4-3* 6 -* 6 -2* 7 }, = l/"-nol 2 *b+3*,+ *io-4* u + 4* u + |/"|» 1i }. It is still necessary to determine the functions $ ( 4 F, y , 2\ and $( 2 F, y. 2 J in order to solve our problem. Since <£ ( 4 F, y , 3 J = $ u , we get by acting on it with the operator (S x — iS y ), the state $(*F, y , 3) : (S. — /S,) *(</>. }, 3) = /3$(^, 1, 3)==(4-/5,)* u , ^(^•I'^^TT {*« + *e + *7}. From the state <P ( 4 F, ~, 3) we get the state &f*F, y. 2J by operating with {L x — iL y ) 254 ANSWKRS AND SOLUTIONS The wave function <T> ( 2 /\ j, s\ is determined by the condition of orthogonality to the three functions 4» i^H, -$ , 3J, <I> ( 2 0, ^ , 3J, $ U Ft .J. , 3) . The normalised function 'I> {^F, i , 3) , which we get in this way is $(«/>, j, 3 ) !=s _^_{_-y'6* 4 + * tt + *6 — 2*,}. Finally we get *( 2/? ' T' 2) = -^- { -2^>8-h*9- ^0 4-/^13} • We have now four states with M s = -g- and ^£ = 2, *(*//, 1,2) = = * {$ 8 — $ 9 -|-3$ 10 — 2* u — SSu + V^,}, = V^w { 2 ^«+ 3 ^« + * io_-4 * 11 ~ h4 * 12 " h ^^"* 13 J' In the same group there are still the two *D states. These two states are mutually orthogonal and also orthogonal to the four states we have just written down. From the condition of orthogonality and normalisation we get the following orthonormal functions : *(a*D, \, 2) = i-{ — * 8 — * 9 + *io+*ia}. }=r{ — 5* 8 +3$ 9 -h*io— 4* n — 35> 12 — 2^6* 18 }. ^84 The other wave functions of the 2 D states corresponding to other values of the two components of the angular momenta are easily ob- tained by applying the operators (L x — iL y ) and(S,» — iS y ). 44. We first of all write down the scheme corresponding to the configuration npn'p. We restrict ourselves to non-negative values of Ms and M L ATOMS 255 M s pp 1 M L 2 ^(1 + , 1 + ) *i(l + . 1") 1 *ai + . o + > %(0 + , 1 + ) $ 6 (1 + , 0") $ 7 (0 + , 1") * 8 (l- f + ) * 9 ((T, 1 + ) *,oO + .-l + )*i 1 <0 + ,0 + ) ^i 2 (-l + , 1 + ) *„(<r, o + ) $ 18 (-r, i + ) From this scheme we see that this configuration corresponds to the following terms: 1 S, *S, X P, 3 P, *D and S D. We shall use as zero-th order wave function the functions given in the table. Since the energy does not depend on the values of M s and M L , the perturbation matrix will be of the form: $2 MfO Mg-0 M S =1 M L -0 Mg-0 M L -0 We shall denote by V the perturbation operator. In the first sub- matrix there is only one 3 D term. Hence 256 ANSWERS AND SOLUTIONS In the second submatrix (M L = 2, M s = 0) we have a combination of two terms, 8 D and 1 D, £ tl) (9D) + £ (1) CD) = ^ 22 + K 33 . In the third submatrix (M L = 1 . M s = l)we have a combination of two terms, 8 Dand 8 P, £<»>(B0)+ £< l) ( 8 P) = V 44 + V B5 . The fourth submatrix (vWz = 1, Af s = 0)we have four terms, 8 D, W, 3 P, *P, In the fifth submatrix (Mi = 0, M s = 1) there are three terms, 8 D, 8 P, and 8 S, £ (1) ( 8 D) + £ (1) ( 3 P) + £ (1) (^) = K 10>1 o + K 11>u + ^ 2 , 12 . Finally in the sixth submatrix (M £ = 0, M s = 0) there are six terms, 3 D, 8 P, «s, tD, »P, and 1 S, £ (1) ( 8 D) + £ (1) (*D) -h £ (1) ( 8 P) + -h JB tl) -OP) + ^ ( 8 5) + ^ 1} C5) = = Vt,, is + V Ut M + V 1B . 15 + V 16 , 16 + V„ , 17 H- V 18 , 18 . From these equations one obtains easily an expression for the term values in terms of the diagonal matrix elements: E {l) eP) = V ii ^ r V b5 -V n , &* ( X P) = V m + V 71 + V 88 + V 99 + V u - V^-V^-V^-V,, and so on. 46. The expression ^{L s -\-2S z )<ge is a small perturbation. We consider Russell -Saunders coupling. In that case H commutes with operators J 2 , J g , L 2 , S 2 . The unperturbed energy levels are charac- terised by the quantum numbers J, L, S. Each of these states is de- generate with respect to the direction of the vector J, the degree of degeneracy being equal to 27+1. Since the matrix elements of the operator (L s -\-2S^ which are ATOMS 257 non-diagonal with respect to J z axe equal to zero, the energy cor- rection is simply equal to the average value of the operator (L g -\- 2S g ) in the state characterised by the quantum numbers J, J g , L, S. To evaluate this average value we substitute in the equation of the prob- lem 29, Section 4, ^ = 1, g 2 ^2, J x = L, J Z = S. The result is -— j$ i3 . 5(S + D-M^ + D ) _j Lg ^ ^* ,/ «l2 + 2/ (J + 1) I J * g - According to what we have said a moment ago we find for the required splitting P (i) eh&e . 51. The energy of an atom in a magnetic field is of the same order of magnitude as the spin-orbit interaction. The spin-orbit interaction operator which is equal to cp(r)/s (see problem 10, Section 7) is of the same order as the operator ^L (f g -\-2s g ) , and V = ?(r)/i+^^(i + 2i) can be considered to be a small perturbation. The square of the z -component of the orbital angular momentum, and also the square of the z -component of the spin, will be constants of motion in the un- perturbed state. It is convenient for us to take another set of con- stants of motion. We shall characterise the stationary unperturbed ptates by the quantum numbers n, I, j, mj (/ 2 , p, j z commute with H ). The degree of the degeneracy in the case of a Coulomb field is equal co 2n 2 , and in the case of arbitrary central field of force 2 (2/ -f- 1). tt is unnecessary for us to solve a secular equation of such a high order. We note that in the perturbed state the square of the orbital ingular momentum and the z -component of the total angular momen- tum are integrals of motion. The wave function of the perturbed problem can therefore be constructed from the functions corresponding to the same value of n, I, m it or, ty = c^o) (», /, j = / — 1 , m,) + c 2 <l>W> (n, I, j = I + \ . »j) Pr r p <b =c. 1 + 258 4-c 2 ANSWERS AND SOLUTIONS R w / y i — m j -+- 2 Y *. ">r v» ^ + 1 \ -/"/+«, + 4- K,. v . = 0. The matrix elements of the operator V axe equal to where °° gfi ^ = J /? nZ (r) 9 (r) /?„! (r) r 2 </r, and Ho = §i^ • The energy value E follows from the solution of the secular equation E° ni + A^ + cWH«ij (l + 27Tl) — E ' We shall denote by E + and E_ the energies of the one-electron atom, taking spin-orbit interaction into account where E + refers to a state ;=/4-V 2 . a 1101 E - to; = / — V*- From problem 10, section 7, it follows that E + = Eni-\-A- 2 -, t_-C.nl * 2 ' The solution of the secular equation is E = y (E + + E_) -f ^J*o«i — Let us consider some limiting cases. a) In the case of weak fields, that is when p $&<CE+ — E_, we get the following expressions for the energy: E = E + -h<^^o^§^T' 21 ATOMS 259 The first energy value corresponds to the energy of the »-th level of the state with j = i-\- i / 2 , and the second to /' = / — V 8 (see problem 46, section 7). b) In the case of strong fields, that is for fi J#^>£ + — E_we have E = 1 (E + + E_) 4- S^h^j ± y J^Vo ± 2^1 ( £ + — £ ->- Let E c be the energy of the centre of gravity of the level when there is no field present, that is, E + (l + l) + E.l (The statistical weights of the states E + and E_ have a ratio / ~*" 1 ) and let AE be the difference £ + — E_. With this notation we get for E E = E c +mH{m i ±\)±4^\{ m i + :} 2)- This expression is identical, as can easily be checked, with the expression of problem 53, section 7. The upper sign corresponds to the state with m l = m i — 7 2 , /» 8 = »/2, and the lower one to the state with m z = w^-|-V2» rn 8 — — 1 /2. 52. c x = y i- (l -f- T ), c 2 = y — (1 — y) for the upper state, Cl = V ¥ ( 1 ~~ t) * c 2 = — K y ( l + T) for the lower state > where 1 ntj ]/ 1 (A£)» + oTTT A£ Wo + X ** 2 Ho 4 v ' ' 2/ + 1 We shall consider some limiting cases. a) Let the magnetic field be vanishingly small, that is AE^> J#|* , with 7 « 1 . We shall get for the upper level c l = l, c 2 = 0, and for the lower level c 1 = 0, c 2 = 1. TO . b) Strong magnetic field AE<C J£> . In that case T = — ^f and we have for the upper level "*" 2~ -1— f 2/4-1 ' 2 ~~ r 2/+1 ' 2 ~~ f 2/+1 and for the lower level 'i— r 2/ + i • C2 r + »»i + y 2/ + 1 ' 2 ~ f 2/-J-1 260 ANSWERS AND SOLUTIONS If we substitute these values of c x and c 2 into the expression for the wave functions (see problem 51, section 7) we get for the upper level ' n ,»-•/, (°» ?)' <!> = *£? and for the lower level 4» = ai3 ^ / 4 53. Since the energy in the magnetic field is considerably larger than the spin-orbit interaction energy, we can neglect the latter to a first approximation. In this case 4 and h are constants of motion and the energy splitting follows from the equation E^= e ±^e{m l + 2m s ). In the second approximation we must take the spin -orbit interaction into account. The multiplet splitting which is added to the splitting in the magnetic field is equal to the average value of the operator -_.-_(/«) (see problem 10, section 7) over the state with given values of m l and m 8 . . For a given value of one of the components of the angular momentum the average value of the other two is equal to zero so that Is == mim a . The energy splitting of the level, when spin -or bit interaction is taken into account, is thus of the form J [t -i -/ i 4 Fig. 39 ■0-4 £ (D_ |^(m l + 2m 8 )-h 2(ji3car3 m l m a We can substitute in this equation the value of -L, expressed in terms of the fine structure splitting when no field is present. One can easily show that (see problem 10, section 7) g 2 ^ 2 J_ _ E n, I, j= 1+ Vi ~~ E n, I, j-l-V, . A£ / + ATOMS 261 We finally get for £W the expression £(1) =ir c ^<" I '+- 2 ' w «)- A£ f WjOT 8 . '+2 In figure 39 we have given the scheme of the splitting of the Is and 2p terms for an alkali metal atom in a strong magnetic field. 54. The perturbation energy in our case is equal to In this expression we shall neglect the last term since the magnetic moment of the nucleus is small compared to the magnetic moment of the electron fa < ~L\ . If we proceed as in problem 51, section 7, we find the following secular equation — c - + £+■ '+T S/fif J$Y {<+$-<*, *■ <9€H i m r = o, where E + and E_ are the energy terms taking hyperfine structure into account, while E + refers to the state with / = /-}- */*, and E_ to f=i — Va, rn t isthe z-component of the total angular moment <m r = /, r 1 , .... — /), and ja = gj^ . Solving the secular equation we find F _£+-(-£_ LE Y*\ 25 + r »r+P ^here AE = E + — E_, and $ = 2ftx&? A£ We can find the order of m agnitude of the m agnetic field strength rtiich produces the splitting from the formula obtained. We have mrned that &? -^- . In the case of the sodium atom we have E.ff<-= 0.0583 cm" 1 = 1.962 x 10" 18 erg; since fi = 0.922 xlO" 20 )e. cm 3 the field strength &6 will be of the order of magnitude of 00 Oe. We consider two limiting cases. a) In the case of weak field, that is when ji #<A£, we find le following expressions for the energy: 262 ANSWERS AND SOLUTIONS £ = E + 4. *?!*.« E = E_ — ^*m f . b) In the case of strong fields, that is when SP^o^AE, we have E = ±(E + +E_)±3Vh- 57. The normalised eigenfunctions of the hydrogen atom in the unperturbed state are of the form (see problem 20, section 4) *». *-•/. /Vj + m i Y i- v.. »«,-vA Co The energy is determined by the two quantum numbers n, j. When a uniform electric field ($ x = t y = 0, S g = $) is present, / z remains a constant of motion, but the orbital angular momentum ceases to be an integral of motion. The matrix elements of the perturbation opera- tor, V=:e$z, for transitions between states with different values of m,j axe equal to zero. The diagonal elements of the operator V axe also equal to zero, that is 2 / "*+zu + dx = 2 ju*_zu_dx. (2) We must therefore evaluate the matrix elements of V corresponding to a transition from the state n, j, nip i = j-\-ij2 to the state n, j, ntj, l = j — 1/2 in order to find the energy splitting. These matrix elements are equal to:- V, = V n = e$ Vf u*-zu + dx = 00 i r B Rn, i-v, (0 «».i+v. 00 <* r 2 V/(7+T) X u X [VU-r-mjiU — mj+l) X X J Kj_ Vt . m rV ,Yi+y» m^-1/.cos dfi - V(y- mjtfj+mf-r- 1 )X X J ^J - v s . »»,•+ v* K /+ Vs. «•_,-+ v, cos » dQ ) . 13) ATOMS 263 We shall first integrate over the angles. Since ., «,, ,n ,.v ,/ (/-+W-f-l)(/ — ffl-i-1) u r „ . , ■ -,/ (/ + III) (I— III) y « v + V (:>/.+ i) (•>/- 1)**- «.»•(•». r) we find that the expression within braces in equation (3) is equal to 2 ^ 7 _^ -{(/ + «fj)(/--/iij+o-(y+^+i)(y--"t J )}== ImCM-D If we then integrate over r, we get We find thus finally for the perturbation matrix elements (3): 3 V^T^¥f , v v We find the required energy correction by solving the secular equa- tion ' i i/i V, 0, e = =tV. =*•/.._(,+!)• (y+1) — For a given value of n the term with j = n — 1 / 2 is not split in the electrical field since it is not degenerate with respect to the quantum number / (/ has a fixed value, l=j — 7 2 =« — 1). All the other terms of the fine structure are split into 2/-f- 1 equidistant levels (m j = —j -fy). IXo=_ 2^yo-+i) m i- 59. In the case under consideration the spin -orbit interaction V v the relativistic correction due to the change in mass, V 2 and the energy of the electron in the external homogeneous electrical field, V 3 = — Fz , are all of the same order of magnitude. We shall there- fore consider their sum as a small perturbation of the original system. For our calculations we shall start from a state where the orbital angular momentum L, its z-component, m l , and the z -component 264 ANSWERS AND SOLUTIONS of the spin, m 8 , (the electrical field is along the 2-axis) have well- defined values. If we evaluate the matrix elements of V v V 2 andK 3 , we have (in atomic units) I'mj mi (m — m{) " v ( /+ t) (/+1) %w for mi = m lt hv w for mi = m -f- m l ~-m — -j or the other way round, otherwise l 3 J HI, 1?! 2 «3 '+4 4* w v»i (V 3 ) lm. 3n ' 2 /(„2_/2 )(/ 2_ 7n 2 ) W=\ F l *> »-» 8 n,;m,' If n = 2 the energy of the state with quantum numbers 1=1, m = m l -\- -f-m 8 = dz 3 / 2 (/ = 8 / a ) will not change in the electrical field. The shift of this level due to V x and V 2 is equal to -^ at. un. (see problem 10, section 7). The splitting of the level with quantum numbers n =2, m =±-^, can be found from the solution of the secular equation, 8/2~ 4 8/2 -U- 7(D 3F 3F 15 s __8. :(») = 0, where 38 = |- at. un. is the fine structure splitting of the level .n=2 when there is no external field. If we introduce in that equation e which is connected with E (1) by the relation E (1) = e — -^8 ( — — 8 4 \ 4 is the energy of the centre of gravity of the three energy levels £i I) +4 1) +4 l) 3 — ?*). we get: — e 8/"2 8/"2 8 — s — 3F — 3F — 8 — 8 = 0, ATOMS 265 or S 3 _ e (38 2 _j_ 9F2) _ 2ga = We shall solve this equation both for weak fields (F<CS) and for strong fields (F»S). m the first case we fin<J ^ > *2 = — S+/3/= > — .^ 8 3 = 28 + 2^. In the second case the result is 1 2 F^ 9 /*' 2 9 /*' 60. The average value of the total energy is equal to Z /*od+*«)&|. (I-f-X«)rf« H= = 7-; .. (1) J Mo V +)*?** If we integrate the enumerator by parts the equation is simplified. I he kinetic energy operator of the electrons is of the form n where « is the number of electrons and A, is the Laplace operator of the i-th electron. (We used atomic units. ) We shall write down an expression for the average value of the kinetic energy in a form which is symmetric in «£ and % , n r= -i|ij{* 1 +^ i i( 1 +^)'Hi(i+^)A i (i+^) t ;j ( / t . If we differentiate under the integral sign we get — n r ^"tUt I (fo(l + X U )2A^ -h i=l * (2) 4- % ( * + W A^o + 2%!»; ( 1 + Xu) A iU + 2X ( 1 + Xu) V, (^ ) v> } d x. We shall transform the last two terms, using the identity = * (I -h *«) A 4 « -+- (1 4- hi) V, (^0) V,« 4- X<& (V iU ) 2 . (3) If we integrate this identity over the whole of configuration space we get: 266 ANSWERS AND SOLUTIONS / {'>o%(l + >") A i" + ( 1 + >-")Vi('T'o'i»o)Vi«M* = =--->. j faVifpdx. (4) Substituting expression (4) into equation (2) we get n ^=- tSy I {*+^> , ^4-%<H-*«> , MJ}* + The Hamiltonian operator // is equal to H = //<,+ « = r-f- V+ «. Since V commutes with (1 +X«), expression (1) can be written as follows: n ■T J (1+Xu)2 *^« )(,, 4 S J ItfMV) 2 * £ = f o+ ? 7~* ~ ' Since // <|» = Eo'V we have n c j iT a (1 + Xo)> *o dx +■ J S J +o *o < V i u ) 3 rfx // = £<H ? ; J <i+kOHo<M x or (b)oo + 2X ( U 2 )oo -j- X2 (a3)oo + y S { (V '" )2 >0 ° tf = £ oH l+2M")oo + X2(" a )w ' <5) where («)„,= f^a-M*. (" 2 )oo = jYo" 2 %<^ and so on. We shall expar the second term in equation (5) in a power series, neglecting terms like (u^oo, (u%> For the energy correction A£ we find an approximate expressio of the form n A£ « («)oo +■ 2X (" 2 )oo - 2^ («& + y £ { (Vi " )2 }o ° ' ^ i = l The variational parameter X is determined from the condition n £A£ ==2(u 2 )oo _2( tt ) 2 <) + Xy{(V i a)Moo = 0, ^ <-i so that we find for X the expression X = 2 (a)So-<A» ATOMS 267 Substituting this value into equation (6) we get the required relation A£«(tf) 00 — 2— . (7) 2 <(V f «) a }oo t=i' 61. If the atom is in a uniform electrical field of strength t>, along the -z-axis the perturbation operator is n i = i whose matrix element («) 00 is equal to zero. From equation (7) of the preceding problem we have AEss — 2& 2 <( * 2)oo} , n where » is the number of electrons. It follows thus that the polarisibility is equal to: a = iil£!)oo>!. n It is necessary to note that this equation which is obtained by using one variational parameter X is a reasonable approximation only for hydrogen and helium. For the case of atoms with several electron shells the deformation of the shells will not be the same. To obtain a better result from the variational method we must thus introduce for each electron shell its own variational parameter. For the hydrogen atom we have oo (* 2 )oo = j (' 2 )oo = T~ J e ~ 2Tri dr = l and thus Q a = 4 at. un. In the cgs system we have For the helium atom we put the ground state wave function in the form (see problem 14, section 7) ^faff^efffV^, where Z eff = g . The final result is a = 0^8 at. un. or a = 0.98 (-^V cm 3 . From this value of a we get for the dielectric constant of helium 268 ANSWERS AND SOLUTIONS under normal conditions s= 1-00049, while its experimental value is 1.00074. The relatively large difference between the theoretical and experimental values is mainly due to the fact that we have used a rather rough approximation for the unperturbed wave function. 62. a)EW B n ."!' '±[ (27 +1) 2 «gg± 8; (; + l) t*c \ v J ± /«/ + »° *• + 3r ( » 2 - (/ + i)'} $2 (See problems 57 and 59, section 7). £ (1) (» = 2, m t = ±|) = «- ~ a 2 (£) . where e follows from the solution of a cubic equation e s + 2fe 2 — s (352 _|_ 9f a _ ^2) ± 282^ — 28 3 = , where >-&*. '-*<£)■ »-&(*)■ For a strong field (8<C^ 8< CP) we have (see problem 59, sec- tion 7) -x— =*=? — »^ — 2-: — 9f — p ' e s = ± ? + 3F + ^ § ^L 82 . 63. Let the z-axis be along a magnetic field direction and the x-axis along the electrical field direction. The potential energy operator of an electron in these fields will then be given by We shall consider this expression to be a small perturbation, and we shall characterise the unperturbed stationary states by the quantum numbers n, I, m, o (m and 3 are the z-components of the orbital angular momentum and the spin). The non -vanishing matrix elements of x axe of the form : ATOMS 269 , „xl-i.». __/~W. '»-!_ 3... , /" (n- — /'-') (/ — w + 1) V — m) w ',™-i- Wj-i. i«— 4 " V (2/+l)(2/— 1) W/_ i, ,„ _ ! — W ; , ,„ — T K (2/-H)(2/-i) ('-£)• We shall consider the case « = 2. Let us fix our ideas and assume that the z-component of the spin is equal to +^. We shall use the notation 3 =^-^, T = JL- e $ a .. In this notation the perturbation J;j.C ' -if 2 operator matrix will be of the form 2-1 — -A 3 t ' -T T ?/ We have numbered the states in order of decreasing values / and m. The state with quantum numbers /= l, m = does not combine with the other states. The energy of that state is equal to E[ l) = p. The other three eigenvalues of the perturbation matrix are determined from the solution of the secular equation E 3 — 33E 2 -f-2(3 2 — T 2 )£+2-,' 2 3 = 0. If we solve it we find E^ = 3, ££ ) 4 =8=tl/"3 2 + 2 T 2 or, using the definitions of /3 and y, 64*. If we used the normalised ground state wave function 6(0 = -^*™ (ass— V „. YV ' Y*a* \ me?) (1) we get for the required average value r*= f r»W(r)]*dV. (2) Using expression (1), integrating over the angles and making the substitution — = t, we get 270 ATOMS ^^.wricv*. in The integral is r(/t-M) so that we get f = r -±±£a». (3) From equations (3) or (2) it is clear that r n has a meaning only for n-> — 3. In particular, we have for n= 1 and » = 2, so that the dispersion of the radius is determined by the equation 65*. Since the protons, which capture the negative mesons, are approximately uniformly distributed over the nuclear volume, we find that the required probability is clearly equal to the ratio *nucl Ji+eoi" dV (1) where t|» is the wave function of the meson in its tf-orbit. The integration in the denominator is over the whole space, and in the enumerator over the nuclear volume V nuc i which is approximately proportional to Z ( V nucl « y*^ 3 = -J ( r o A ' f 'f = -y- r l A » -^-rl-2Z, where /-Q^const^ 1.2- 10" 13 cm). We shall assume that the nuclear radius R is small compared to the radius of the /("-orbit of the meson a = -g-^- , where \k is the meson mass.. Taking into account the fact that the mesonic wave function varies appreciably only over distances of the order of magni tude of a, we get for the enumerator ?« | ty (0) | a • V nuc i, where \J/(0) is the value of the mesonic wave function at the origin (that is, in- side the nucleus). If we use the normalised wave function (which makes the de- nominator equal to unity), ATOMS 271 it is immediately clear (since a — -~- ) that the ratio (1), and thus also the required probability, is proportional to Z 8 Z = Z*. We note that our approximation a 7$> R is not valid for heavy nuclei. Indeed, since /?«/yi ,/, ;^r (2Z) ,/ ^ where r o^^[jtff) a t. un. and ix«200at. un. the inequality a ^> R is equivalent to (compare problem 15, section 5), Z <C 45. 66*. The potential 9 to be derived is the sum of the potential <£ p = — , due to the proton and the potential ~y t , due to the electron (the averaging is over the wavef unction of the hydrogen atom ground state %== T^ e a ' with a== -^ h <pC) = <Pp(r)-f f e (r) = -^-e \^£^dV'. (1) The potential cp e , is, of course, the same as the potential of a static electron "cloud" of density p(r')z= — «<$ (/•'). We can thus simply find ~cp e as the spherically symmetric solution of Poisson's equation I*^__4, W _te 1 ^.-T. from which i a is easily obtained by integrating twice from r to 00. The integral in equation (1) is most easily evaluated by using f° r 1 r _ r / 1 an expansion in Legendre functions 00 1 « r'<r ^- 7T = I2/',(cos9')(4). if r>>r ^^=J ? |; Pl (cos6')(f)'. J-0 where 8' is the angle between r' and r. If we split the integration over r' into two parts (r' < r and r' >- r). we see that in both parts integration over the angle 8' reduces the <x> sums 2 to their first term ( / = 0) by virtue of the orthogonality 272 ANSWERS AND SOLUTIONS of the Legendre functions, and we get As was clear from the start, the potential £ (and thus also <p) is spherically symmetric. If we perform the integration we find finally tW _.(i+i).— . In the limits r <C a and r ^> a we get the translucent result 2r 9 « — (Coulomb field of the proton) and <p » — e~ « (proton prac- tically completely screened by the electron). 67.* The average value of the dipole moment of a system of N particles is equal to <</) = j . . . J f (r x , .... r^S ViJ'K'i 'jrVV"**. where e ( is the charge of the i-th particle 2 *i r i = <* tne dipole mo ~ ment operator, which in the coordinate representation is simply equivalent to multiplying by d « 2 ** r < and the integration is over the configuration space of all particles. If the system in a state of well defined parity, <|j*<1» is an even function in the coordinates of all the particles, while the dipole moment d is odd in those coordi- nates; that is, the complete expression under the integral sign is thus odd. Since the integration goes to infinity for each of the co- ordinates, that is, is over a symmetrical domain, it is clear that (rf) = 0. This result can be obtained also in a different (less rigorous) way. Since the reflection operator commutes with the operators of the angular momentum components and the latter commute with one another in the semi-classical case, that is, if Af 4 ^>fi any state with a well-defined parity has a well-defined value for the angular mom- entum vector J (in particular, the spin of the system I). A non-vanishing angular momentum defines a definite direction in space, which must also be the direction of the (average) dipole moment (d): y (d) = const J. (1) Since {d) is a polar vector and / an axial vector, the (physically necessary) invariance of equation (1) under a reflection means that ATOMS 273 the constant in equation (1) must be a pseudoscalar (which in our case is clearly not so). The constant must thus be equal to zero, or, (d) = 0. 68. * The operator of the ^-component of the dipole moment of a system of particles of mass y. and charge e is of the form N 4» = *2 x i> and its matrix elements d mn , which occur in the sum N •2 rule are equal to where ¥ m , *„ , . . . form a complete orthonormal set of station- ary wave functions for the system of particles under consideration (without time factors). We shall introduce the operator of the time derivative of the dipole moment component d (we drop the index * from now on) and use the formula A — / En ~ E m d u nm — * g u nm' where E n are the energy levels of the system of particles; we find easily the following chain of equations, ^(E n -E m )\d mn \^^^(E n -E m )d nm d mn = n n = "2 T ^ "nm"mn ~2~[ 2j ^n^nm == "2 "J" v"* ""/mm* n n we have used the fact that d is a Hermitean operator (d mn = d nm ) and the multiplication rule for matrices. If we take the definition of d into account and the fact that the coordinate operators x i commute with the velocity operators **-•£■** if '¥■*. while we find N 2 ( £ n — ^m) I dmn |* = y T '* 2 ( * '*« ~~ ****>»»»» = 2JT ^' which had to be proved . 274 ANSWERS AND SOLUTIONS We emphasise that the sum is independent of m, that is, it is the same for all "initial" states. 69. * We must add to the Hamiltonian of the unperturbed problem (see problem 23, section 1) the perturbation energy operator #' = — rf«8 = — d-$cos<p, so that the Schrodinger equation will be of the form where <p is the angle of rotation, and E the energy of the rotator. The energy levels of the unperturbed problem, are twofold degenerate, apart from the lowest level, with respect to the direction of the angular momentum M g = mh. However, since this degeneracy is not lifted until we take the second order perturbation theory into account we can up to that order apply the theory of perturbation of non-degenerate levels. (The level with m = ± 1 which is split, as was shown by Trubnikow, must be ex- cluded). In other words the unperturbed wave functions d>(0) = -■]— e im * are still good functions for the zero-th approximation. The perturbation operator matrix elements are equal to 2k 2 « [ if m'j=m±l, = — ^| J e i (»'-«»)* cos cpdcp = ! _d$ if m / _„-£!. It follows from this that the first order energy correction of all levels is equal to zero. The second-order correction of the m-th level is clearly given by c (2) I H 'm, «t-l I 2 , \ H m,m + l \* Id t E™-E^_ x ^ E%-E% +1 ft 2 (4m 2 - 1) so that the energy levels of the plane rotator in a weak electrical field are up to second order terms given by P F «» . F W . E (2>-^-4- f<P & . t m = Cm + c i»T c i» — 2/ ' tfi (4m a — 1) ATOMS 275 This result can be interpreted very simply. If we introduce in the usual manner the polarisibility, a , of the rotator as the ratio of the induced dipole moment to the external field strength, we find that the energy of the induced dipole in the field £ , which 6 is equal to the work of polarisation, — a f $dt) is equal to -^ aS 2 . If we compare this additional energy with the first non-vanishing energy correction, which we have just obtained, we find the follow- ing formula for the polarisation, 2A/2 a m = fi2 (4 TO 2_i)' We see that for m =£ , the polarisibility is negative, that is, the electrical moment of the rotator, <f, is oriented antiparallel to the field 8. The opposite is true for m = 0. This result is completely in accordance with the classical effect of anti -parallel polarisation for fast rotation of a plane rotator, and parallel polarisation for slow rotation. Fast is meant here to be sufficiently fast for the rotator to be completely turned over in the field. 70. * We shall choose the direction of 8 as the polar axis of a spherical system of coordinates, so that the perturbation operator will be y;=v = — d8 =— d-gcose, (i) where is the angle between the axis of the rotator (which is parallel to d) and the field 8. The wave functions of the unperturbed stationary states of the rotator are of the form ♦8(8. ?)- *W8, T )- ]f ^El^ PT(co^)e im \ (2) where 6 is the polar angle introduced a moment ago, and 9 the azimuthal angle of the axis of the rotator. The functions (2) are orthonormal, J J Y* m (6, cp) Y lm > (0, cp) sin 8 d8 dcp = Mm»« • (3) In particular, the ground state wave function for whose pertur- bation we are looking, is of the form ♦BO.,)-!'.-^. (4) 276 ANSWERS AND SOLUTIONS The unperturbed energy levels of the stationary states are ^1 to iff, '<* + '>» , EP-o. (5) We shall write down the perturbation operator matrix element for a transition from the ground state ( 0, 0) to the excited state (l> m), VS-CVSD* — / Jlt8(«. f) jVtJ.fi (9. <p)sin6d6d ? . If we use equations (1), (2), (4), the orthonormality condition (3), and replace cosO by y r ^r 10 (». ?) we get <V& ^V^^-d.s/jX d-$ x 1 J f r; w (0, cp)F 10 (9, cp)sin6d8dc P = ^^|8 I1 8«o- (6) There is thus only one non-vanishing matrix element, namely, the one from the ground state 1= ™ = to the state / = 1, m = 0. Since this matrix element is non-diagonal the first order energy level correction vanishes. The second order correction is according to the general equa- tions of perturbation theory and using equations (5) and (6) equal to to — Zu Zt fig»)_M0) ~ 4»-£<°> 3ft 2 * K) The shift of the ground state level of a three-dimensional rotator is thus negative; the same would be true for the second order ground state energy correction of any quantum mechanical system . The condition of applicability of the perturbation theory is that the shift (7) is small compared to the difference |^ 0) — M 0) l between the ground state and the first excited state. If we solve this condition for $, we get ATOMS 277 As we should have expected this inequality also means that the perturbation energy ( ~$<t ) is small compared to the distance between consecutive levels ( — fi 2 //). Finally we note that we can use recurrence relations of the form cos 6 • Y lm = a, w Kj_ lf „, + b lm Y ln> m to solve very easily the problem of the perturbation of excited state of the rotator with arbitrary values of I and m. 71. * We shall denote the radius vectors of the electrons with respect to the nucleus by r t and r 2 . We have the following SchrSdinger equation for the stationary states of the helium -like two-electron system, H^(r v r 2 )==EW(r 1 , r 2 ), (1) where the Hamiltonian H is of the form We consider the last term as a perturbation, and denote it by H'. We shall perform our calculations in atomic units (fi= 1, e= 1, m=l). Since the unperturbed Hamiltonian, H m = H—tf is the sum of two one -electron Hamiltonians, the unperturbed wave function of the ground state of our system will be the product of two hydrogen - like functions in the field of the nucleus Z (since the ground state of hydrogen -like atom lies far below the excited states, it is clear that the ground state of a helium -like system can be constructed out of the ground states of the corresponding hydrogen -like systems), and the unperturbed ground state energy is equal to the sum of the ground state energy of two hydrogen -like atoms, ,(0) Z2 Za_ -2 278 ANSWERS AND SOLUTIONS The required first-order correction to the ground state energy is equal to the average value of the perturbing energy H' in the unperturbed state (3), £(!)_//_ j" §tf (r u r 2 )tfV (r lt r 2 )dr t dr 2 a %[ \ e -2Z(r l+ rj±_ drid r 2 . it 2 J J r& (5) The integral we have just obtained can be evaluated in different ways. We shall use one of the more simple ways. The quantity _L — _JL__ is the generating function for the r 12 \ri — r 2 1 Legendre polynomials, oo = < |r t — r 2 oo rX(*) ,p|(C0,6) if ri</ " 8 ' < 6 "> 2 J=0 where 6 is the angle between r 4 and r 2 . If we integrate, for instance, first over r 2 and split the domain of integration accord- ing to equations (6 f ) and (6 n ) we get: If we take the polar axis of a spherical system of coordinates along r t and use the orthogonality relation of the Legend^ poly- nomials, it « fp,(cos6)sin8d8== J Pi(cose)P (cos6)siti8rf8 =^^-810, we find that after integrating over the angle only two term s are left in the expression within braces, ATOMS 279 If we now perform the remaining elementary integration we find finally, ^-JZ. (7) The positive sign of this correction corresponds, as should be, to the mutual repulsion of the electrons. From equations (4) and (7) we get for the ground state energy of a helium -like atom or ion to a first approximation, E^E m + E (l) = -(z>~. §z). or in the usual units, (8) ^_(z>-Az)^=-_(z*_4z)27.1eV. (8,) The single ionisation potential V t is the energy necessary to take one of the electrons just into a state of the continuous spectrum. It is clear that this quantity is equal to the difference between the total ionisation energy, -£„ given by equation (8), and the ener- gy of ionisation of a one -electron atom or ion, which is equal to 2Z 2 at. un. , K 1 = (z 2 -4z)_^Z 2 = (4z 2 -Az)at.un.= -(lz"_|z) 27.1 eV. Perturbation theory is applicable to the present problem pro- vided Z^> 1, as can be seen from equation (8). (In the given simple example we do not have to consider very strictly the re- quirement of the applicability of perturbation theory; this is ne- cessary if we want to find the first -order correction to the wave function or the second order correction to the energy. ) It is of interest to note, however, that our evaluation, which is relatively gross, leads already for Z=-. 2 (helium atom) to a value of the binding energy which agrees with the experimental value within about 5%. For larger values of Z ( Z = 3: Li + , Z = 4: Be , — ) the agreement is naturally even better. 280 ANSWERS AND SOLUTIONS The agreement with experimental values is not so good for the first ionisation potential V l evaluated from equation (9); for helium, for instance (7 = 2), the difference is about 15%. 72. * First of all we show that the characteristic time para- meter r enters into the perturbing field in such a way that the total pulse /' which is classically transferred to the oscillator by the electrical field over the total duration of the perturbation, does not depend on t , dt — eA= const. (1) This means graphically that the area under the curve is the same for all values of r (see fig. 40). The probability for a transition from the n-th stationary state of the discrete spectrum to the k-th. is equal to ««* = CO I V kn e u *rS d t (2) CO where V kn = ( tyt'vtfn dx is the matrix element of the perturba- tion V, «. tel = |[^-e!? ) ]. where tf\ $>. E<£\ fi? are the wave functions and energy levels of the corresponding (unperturbed) stationary states. If we denote by e, jj., and u> the charge, the mass, and the eigenfrequency of the oscillator, and by x its displacement from its equilibrium position we get in the case under consideration of £ uniform field for the perturbation operator, V(x, t) = — ex&(t)~x. ATOMS 281 It is well known that in the matrix of the coordinate of the oscillator in the energy representation only the following matrix elements are different from zero, x n> n+1 = x n+li n =y (n j~^ fi . Since we have assumed that the oscillator was original in its ground state ( n = ), we are dealing with only the following non- vanishing matrix elements of the perturbation: K M = ^ = --^/-X e -(4)". (3) In first -order perturbation theory the uniform field can thus produce a position of the oscillator only to the first excited state ( &= n + 1 = 1). (To evaluate the probability for a transition to the second excited state we must use second order perturbation theory - for this transition, n = 0-*>n = 2 , goes through the intermediate state n = l and so on; each excited energy state corresponds to a certain order of perturbation theory and a correspondingly in- creasing number of intermediate states. From the character of the coordinate matrix it follows that each of the intermediate states is next to the previous one - for instance, -* 1 -*2 -»3 • ) If we substitute expression (3) and also u) te = a) loE i . h . [£l 0) — E ( o ] ] = o) into equation (2) we get the probability for excitation, pa oo Vt/, 01 2irc2(Afi If we evaluate the well-known integral oo \e i ? x -** , dx = ]/~^ — oo we get finally dt (4) "~— ^5T« pi _!(«,)» (5) We see that for a given total classical transferred pulse P the probability for excitation decreases steeply with increasing effective duration of the perturbation x and if -c ^> — (duration 282 ANSWERS AND SOLUTIONS of perturbation much less than the classical period of the oscilla- tor), this probability is very small; and we are dealing with a so- called adiabatic perturbation. On the other hand for a fast perturbation ( x <C - ) the proba- bility of excitation (5) is practically constant. In the limit where t -» we can use a well-known expression for the 6 -function, Hmg(0 = i48(0 = — 8(f) ; we are dealing with a so-called instantaneous (or ballistic) pertur- bation and the probability will be equal to Ifo'-JiK' _ < 5 '> which is equal to the ratio of the classical imparted energy, ^-P 9 , to the distance between the oscillator energy levels Aw. The criterion of applicability of perturbation theory is that the probability for excitation should be small compared to the proba- bility that the oscillator stays in the ground state, w i *C (1 — t»oi>» or %<1. (6) We see from equation (5) that a sufficient condition for satis- fying conditions (6) is that It is clear, however, that if the degree of adiabatic change is sufficiently large, that is, if r » aT 1 , condition (6') is much too strong, and perturbation theory is also applicable if ^-P 2 is of the order of fi<o. We emphasize that the excitation of the oscillator when conditioi (6») is satisfied is a quantum effect. Indeed, in classical mechan- ics excitation of the oscillator by an amount h<o would in general not be possible because of the conservation law for the energy. In quantum mechanics, however, the excitation is possible even though we can see from equations (5) and (5 1 ) that its probability is small. This does not violate the law of conservation of energy, since the quantity —■ in quantum mechanics cannot, in general, be interpreted in the 'classical way as the energy given to the oscil- lator by the field. 73. * The perturbation in the present case is of a form which is qualitatively the same as the Gaussian perturbation of the pre- ceding problem. The condition P — f e% if) dt =» const leads, as ATOMS 283 can easily be verified, to the following law of change of the field «» = T^ (1) We can use all equations of the preceding problem without any change up to equation (3) which now will be V n so that we get for the probability of a transition to the first excited state instead of equation (4) „> /*e« I f **»' dt f — oo The integral in expression (3) can be most simply evaluated using the theory of residues. We use a variable t in the complex plane and we close the circuit, which starts along the real axis corres- ponding to the integration in expression (3), by a semi -circle of infinitely large radius in the upper half plane; the result of inte- gration over this contour is the same as the result of the integral in expression (3), since the integrand contains a factor e iu>t . (It is of course necessary that the rest of the integrand also tends to zero if |*|->oo). The only singularity (in our case a pole of the first order) of the integrand inside the contour is at the point t = -\-tz. The integral over t is equal to 2*/ times the residue at that point, gfa>* dt = 2tt/ Res gtatt fl + X 2 = 2ir/ 2fr Substituting this result into equation (3) we find finally: w oi = ^'- 2u,T - (4) The qualitative dependence of this transition probability on the effect of duration of the perturbation t and also the discussion of the applicability of perturbation theory can proceed exactly as in the preceding problem. We note that in the limiting case of an instan- taneous perturbation (t -* 0) it follows from equation (1) and one 284 ANSWERS AND SOLUTIONS possible expression for the <5 -function lim $ (t) = — P3 (t), that the probability of excitation tends to 2 . -, which is exactly the same as the result (5') of the preceding problem. We should have ex- pected this result since for a given value of f &(t)dt the transi- — oo tion under the action of a sudden perturbation should not depend on the concrete form of the "pulse" function §(/). This can be formally expressed also by noting that there are a number of dif- ferent "pulse" functions which lead all to the same function 8(t). 74. * The general equation for the probability of a transition per unit time from a state of the discrete spectrum into a state corresponding to an infinitesimal energy interval of the continuous spectrum through the action of a periodic perturbation of frequency oj is of the form dw m = 1 1 F m I* 8 (E v - Z4 0) - M dv. (1) In this expression n is a set of quantum numbers characterising the state of the discrete spectrum, v a set of quantities for the states of the continuous spectrum , rfv the corresponding infinit- esimal interval, E® and f v are the unperturbed energy levels of the discrete spectrum and the continuous spectrum, while F Hn is the matrix element of the perturbation operator for the transition under consideration, which will be defined more exactly in a moment. The wave functions of the discrete spectrum are norma- lised to unity and those of the continuous spectrum to 8(v — /). Let r and-e be the radius vector and charge of the atomic el- ectron and let 8 be the amplitude of the intensity of the uniform electrical field. In our case the perturbation operator V is of the form V = e&(t)r = eg ( fsinti>t = Fe- iu > t -\-F'e to > t , (2) where » , / 7 = 2^V- (2») The 6 -function which occurs in equation (1) shows the reson- ance character of possible transitions; the energy can make a transition to a state of the continuous spectrum only if (strictly speaking, equation (3) is only approximately considered as we shall discuss at the end of this problem) £ v = £f-Mco (3) (absorption of a "quantum" of frequency co). Since for the hydrogen atom in the ground state (£ v — £« 0) ) mi n = S» iz f oUo ws that the ATOMS 285 minimum frequency which is necessary for ionisation is given by the equation i We shall now evaluate the matrix element F;n=j'XF<WdT. (4) We shall substitute in this expression equation (2 ? ) and also ^ = <Woo = (™*y V 'e- r/a , ->., » % = (2*)-\ ikr , where we have used the normalisation mentioned earlier. We shall choose for our set of quantum numbers v the wave vector of the free electron k\ it is clear that the use of plane waves for ip v is, strictly speaking, only correct if w ^> <b , that is, when the electron is moving fast. We get in this way from equation (4) F vn = ^.(2i:)- v *(*a»r ,A J e-**'-'-A'(g r) dr. (4») To evaluate the integral we introduce a spherical system of coordinates (r, 0, <j>) with the polar axis along k and we denote by 8 the angle between k and 8 . We get then %>(f = $(/ [cos 6 cos" -f - sin B sin cos (cp — <p )], where cp is the azimuth of 8 in this system of coordinates. Sub- stituting this last expression into equation (4') the second term will clearly give zero, when integrated over 9, and we get (cosOsx): /?vn = 2 i ^~^ 2ltg ° COSe J I J e ~ XhrX a ^ dr ) xdx = _ ie$ cos 6 C 3!xdx __ ej, cos 8 Wka* ~ * (2a) 3 /« J (1 + /**)* ~~ »( 2 ») ,/ ' < l + ****• Substituting this expression for \F m \ z into equation (1) and writing di = dk = k*dk dQ h = &■—- dQ k dE, = —. d Q h dE, t,2b2 ' (we use the equation £-, = -5^-). we get: 96 ma V $*k 3 cos 2 9 ^ tm dw «' -n'-W iT+J^HE.-EV-h^dQKdE.,, where dQ h is an element of solid angle with its axis along k. If we finally integrate dw ni over E,, we find the probability of ioni- sation while the electron has a final wave vector k within the ele- ment d&k . For this integration we need clearly only consider the point E, = £<?> -f. hw, that is & = ~£ = ^ (to — u) ), so that (1 -4- kW) = - (we have put Ef « — ~ = — fio> J. I f we also 286 ANSWERS AND SOLUTIONS take into account the fact that a = — ^ , we find finally The angular distribution of the electrons which are expelled from the atom by a high frequency electrical field have axial sym - metry around the only preferred direction, namely, that of the field 8 , as we should have expected since the initial state of the atom was spherically symmetric. The angular distribution is also symmetrical with respect to the plane = ^-, where it is equal to zero (no electrons fly away in a direction perpendicular to the field 8); in other words forward (0 p^ 0) , and backward (0 » rc) expulsion have the same probability. This last result is also natural because the field 8(0 is oscil- lating with a frequency which is much larger than the eigenfre- quency u> of the electron in the atom. It is useful, however, to note that this result (which is qualitatively clear also in classical mechanics) follows automatically from the quantum mechanical calculation. The appearance in expression (5) of the factor cos 2 0, which is symmetric with respect to the directions 8 and — 8 . is caused simply by the structure of the matrix element F m which in general does not depend on the time and which thus cannot be connected with a field of sufficiently high frequency. In actual fact, however, this high frequency behaviour is contained in the energy 6-function, and for frequencies of the field which are less than <d , the probability (1) is strictly equal to zero. (One can see in other words that if »<a>o the 6-function in expression (1) does in general exclude the appearance of the matrix element (4') with real values of k which means that the transition of an electron to the continuous spectrum is impossible in that case). If we integrate expression (5) over all possible angles, and use the fact that 4u cos a «= -£ we get the total probability ">j of the ioni- sation of an atom per unit time as a function of the frequency of the field ««<<•>- -3- if M -As;-- 1 ] • (6) If o)^,o) , that is, if we are near the threshold for ionisation this probability increases from zero as (co — w )\ If <o^>o) it de- creases steeply as <a~ a/t , with increasing co. By differentiating expression (6) one sees easily that it has a maximum for a> = -£U) c . In conclusion we want to discuss briefly the limit of applicabi- lity of the perturbation theory which we have used. First of all it follows from equation (1) that the time t reckoned from * = ATOMS 287 must be sufficiently large. The uncertainty A£ v of the energy of the perturbed final states of the electron is connected to the value of t by the uncertainty relation A£., • t — h. On the other hand the time t cannot be too large since the dis- tortion of the initial wave function of the atom must still be rela- tively small. This is equivalent to requiring that for all interme- diate times t the probability for ionisation, which is equal to w.t, must be less than unity. One can use equation (6) to obtain the limits of applicability of the perturbation theory when we change one of the three quantities /, & , or co while keeping the other two fixed. 75. * The wave function of the initial state of the electron, in atomic units, is of the form y 7i The final state of the electron can be described by a plane wave since its velocity is large, T1 (2*)'/« The perturbing interaction, also in atomic units , is of the form Lt\r~r p \' p where the sum extends over all protons in the nucleus. We shall denote by *F and W v the initial and final states of the nucleus, so that we can write the matrix element of the per- turbation in the form Vm= — I dVdxW*,—^—- e~ ikr \ 1 w IlLo-Zt / n 01 J J \2t^ Zl\r-r p \^y~ e . (1) where dr is the nuclear volume element, and dV the electron volume element. We shall expand the potential of the Coulomb interaction in a Fourier integral, p p p If we substitute this expansion into expression (1) and change the order of integration, we get p From equation (2) it follows that ? e ff r '~ 1. Since moreover r^>r p (the dimensions of the K -orbit are much larger than the dimensions of the nucleus) we have \qr p \<^\. We can thus expand the exponent 288 ANSWERS AND SOLUTIONS in the integral in powers of ( qr p ) and discard all terms from the third onwards, J p • J p = -iq{dx Sfi 2 r,% = - tqd ov j p where we have used the mutual orthogonality of the nuclear wave functions ^¥* and W and where we have denoted by d ol the matrix element of the nuclear dipole moment. If we integrate over the electron coordinates, we get ) dVeKq = [Z * + (q-m* We get thus _ v _,?V2 . [dq q "01 — ' ~^3 "oi J q 2 [z* + (q — kf\*' In the integral over q we introduce a new variable q' = q — k; f q' + k dq' The integrand has a maximum near q f = 0. The width of this maximum is of the order Z Since the final velocity of the elec- tron is much larger than its velocity in the atom, we have k ^> Z. We can to a first approximation neglect therefore the terms in q' ', and we find, k f dq' ifl k A 2 J (Z a +?' 2 ) a Zk*' We get thus finally The probability that the electron leaves within a solid angle rfQ is equal to the following expression: where we have used the general formula for the transition proba- bility under the action of a constant perturbation. If we integrate this expression over all angles, we get the to- tal probability of ejecting an electron per unit time «=3- T |4,i| 2 . or using ordinary units, _16m3£8Z%^j . I2 W ~ 3 fi? flt, I 4n i • On the other hand the probability for dipole radiation is where u is tne frequency of the radiation. We find thus that the internal conversion coefficient, which is ATOMS 289 defined as w a _ w w rad + «'rad is equal to a _ \he ) bv\hm ) 76. * The first inequality shows that the wave function of the atom, that is of the electron shell, does not have time to change appreciably during the duration of the kick t. The second inequa- lity shows that the nucleus can be assumed to have stayed practi - cally fixed during the kick. To find the required transition proba- bility we must expand the original wave function of the electron shell <^ (see first condition! ) in terms of the eigenf unctions of the shell with respect to the moving nucleus; each of these functions describes a stationary state of the moving atom. (The set of these functions includes, of course, also the functions corresponding to the continuous spectrum, that is, describing states of the ionised atom. ) The coefficients of this expansion determine also accord- ing to the general rules of quantum mechanics the probability that the atom is in the corresponding states. We can obtain this expansion either in a system of coordinates in which the nucleus is originally at rest or, more conveniently, in a system of coordinates which is moving with the nucleus after the kick. In this last system the eigenfunctions of the possible final states are the set of the ordinary stationary wave functions of an atom at rest* ^n(fi, r 2 , .... r it ..., r N ), (1) where r i is the coordinate of the t -th electron of the shell with res- pect to the nucleus, and where the index n denotes the totality of quantum numbers characterising the stationary states of the atom. The initial wave function can be transformed if we go over to our present system of coordinates, to the form imv 2 r t i *o = * * %( r v r 2 r it ..., r N ), < 2 > where m is the mass of an electron. 290 ANSWERS AND SOLUTIONS Indeed, the exponential factor is the wave function of the centre of gravity of the electron shell which in our chosen system of coordinates moves clearly with a velocity -v while ^(^ r N ) is the wave function of the shell in its own centre of mass system. (We have ^ j 5.^ *c.o.nu= ex P { J P co.m. R cwn. ] " ex P j { 2 mii ~ V) J^\' from which we get the required expression for e^ r because «! = ma =* . . .=m < =...=m. ) Because of our second condition the coordinates r i in expression (2) can be taken from the same origin as in expression (1). The required expansion of ^ is of the form #>(r lt .... r 4 r N ) = 2 c n ^ n (r v .... r* r N ). n If we multiply both sides of this equation by <!>* dxi ... dii . . . d* N (dx t = dx t dy { dz t ) and integrate over all electrons coordinates, and sum over all their spins, we get, assuming the orthonormality of the functions *n» 1to • • • c « = S - • • / ^ (r i r < r *> x X * * fy>( r i» • • • » r i r*)te\ ... dii ... d%H. The required probability for a transition to the state n is equal to <*«=k»l 2 = N (3) where we have introduced the notation N «=1 We can easily generalise our result to the case of a transition to the continuous spectrum, that is, to the case of ionisation of the atom. We note that provided qa <^ \ the probability (3) to excite an atom "by a kick" is proportional to the probability for the corres- ponding optical transition . Indeed, in that case one can expand the exponential factor and retain only the first two terms, ATOMS 291 The first term gives zero since % and <J> n are orthogonal to one another. If we choose the z-axis in the direction of q we get, which is proportional to the probability for the optical transition 77. * According to equation (3) of the preceding problem the probability for a transition of the hydrogen atom to the n-th sta- tionary state is equal to •w -| /*•'■'**■ f. (1) where % _ («,«)- V'" ( a = J^ is the first Bohr radius), and * T I v ls the r ecoil velocity of the proton). Since «> = -£■, where AT is the proton mass, we have „ m p qs — unr- (2) The required total probability of excitation and ionisation of the atom is clearly equal to 1 — w , where w is the probability that the atom stays in the ground state which according to equation (1) is equal to:- ^o = |/to*''* r dr| a , which after substitution for <p 8 leads to "°~m.\j<-^«rf. (3, The integral occurring in this equation has been evaluated before (see problem 75, section 7); it is equal to -^ iqr dr~ ma * (4 + The probability that the atom stays in the ground state is thus equal to l 7 1 \V (l+-yflW) (4) and the total probability for excitation and ionisation of the atom is equal to 1 — •o— 1— ? f rj. (5) 292 ANSWERS AND SOLUTIONS As should be expected, in the limited cases of weak f-g-^a <C 1 ) and strong (^ga^> l) kicks the probability (5) tends respect- ively to zero or unity. The corresponding limiting expressions are of the form j We note that these results can be seen qualitatively without expli- cit calculations from equation (3) for w : this quantity is small or nearly equal to unity according to whether or not the factor e** r oscillates fast over the range of the factor e~ Zr i a , and the measure of whether the oscillation is fast is the quantity 4p We formulated in the preceding problem the criterion for the applicability of the approximation we have used. Since the ionisa- tion of the atom plays an essential role in the problem under consi- deration we must slightly alter the first part of the criterion, namely, the criterion of the suddenness (ballisticity) of the impact. The length of the impact t must, namely, be small compared to the Bohr period of the most important transitions, where E and E & are the energies of the initial and the final state. In our case E = ^p- , and E k { s equal to the kinetic energy of the electron which is flying away. The effective value of E k can be estimated from equation (1). Since for ga ^> i the integral in equation (1) is appreciably different from zero only for those states whose wave functions, 01> W ) contain a factor of the kind e ikr with k « q , because only under those circumstances can the fast oscillating factors in the function under the integral sign cancel each other's effect. In that case we have thus (£ * } eff m -M E P' (?) where E p is the recoil energy of the proton. If ga <C 1 the value of the integral in equation (1) is effective- ly determined, as can easily be understood, by the degree of over- lap of the functions t^ and «|£, so that an appreciable transition probability is obtained only to final states with k <.— , so that fi a a ( E *\li ~m"^~ E o (8) ATOMS 293 Either equations (6) and (7) or equations (6) and (8), depending on the value of qa, that is, of P , give thus the required condition for the applicability of the "ballistic" approximation. 78. * To derive the quantum rule we want it is necessary to proceed in the same way as in the derivation of the usual Bohr- Sommerfeld quantisation rule. Especially it is necessary to find the exact functions near the points where the semi-classical appro- ximation breaks down. Comparing these functions with the semi- classical ones in the "overlap" region (that is, in the region where both the exact and the semi-classical functions can both be applied) we can find the quantisation rule. It is easily understood that in the semi-classical case which interests us (n ^> I) the exact functions are Coulomb functions for n -► oo(E -» 0), and the overlap region is the interval Z' 1 <Cr<^i Z~' h (we use atomic units). Indeed, the self -consistent field ?(/-), in which each electron of the Thorn as -Fermi distribution moves agrees for distances r <C Z' % with the Coulomb field of the nucleus ?( r )^7» The semi -classical approximation can be applied for distances r^>z~ l . The Coulomb functions for / = 0, ».-> oo (that is, E -» ) are of the form *t = c h( ^l r) (in Coulomb units) or ^ = z 3 /'c y i(>^) ( in atomic units) (1) For r^>Z \ or Zr^>l, we can use the asymptotic expression for the Bessel function J v which is - (»"*-?) ; *'"(^-T) (2) The semi-classical function can be written in the form sin fp r dr + a\ Tsemicl c semicl r yjj~ ' 294 ANSWERS' AND SOLUTIONS Our problem is to determine the phases a. In the region which interests us, the radial momentum p r =» Yl [E— £/(/■)] (for the case I = which we are considering there is no centrifugal potential) is equal to ]/ --, since \U(r)\~ j~^>\E\; the last inequality is true by virtue of the fact that the energy of an electron in the Thorn as -Fermi distribution is of the order Z v \ We have thus r r o o and the wave function is equal to ^ sin (/8Zr+ a) , s\n (Y$Zr-\- «) ji c ] ' semicl /2Z\^ semicl f .*/ l r . r n Comparing the two functions we find that a = — ^ . The semi- classical function is thus of the form Tsemicl c semicl (f sin I | p r dr ■ (4) r YPr On the other hand, the function satisfying the boundary condition near the turning point r , which limits the motion of the particle at large values of r , is of the form sin (/»*+*) Tsemicl ts= C seraicl r YfT ' In order that expressions (4) and (5) are identical it is necessary that the sum of the phases of the sines is equal to a multiple of ir which gives us the quantisation rule we are looking for )p r dr = mt for J* /2j* [E — U (r)] dr = nhn. (6) o o The quantisation rule we have obtained is thus different from the Bohr -Somm erf eld quantisation rule. One sees easily that in the particular case of a Coulomb field Ze* U(r) = — in the whole of space the quantisation rule (6) leads ATOMS 295 to the usual form of the spectrum F Z'pci 1 (This can be checked by substituting into equation (6) for r from the equation E — U (r ) = E + — = , or r = — — , and r o E performing the elementary integration. ) 79. * When the applied electrical field is of the same order as the Thomas-Fermi field, that is, the electrical field in the atom in the region where there are electrons of the Thorn as -Fermi distri- bution, the relative change in the distribution will be of the order of unity. The dipole moment of the atom will thus in that case be of the order of the Thomas -Fermi (TF) value: Thomas -Fermi radius for one electron, and Zr rF for the whole atom. If we assume that the dipole moment is proportional to the field, we get for any, not too large, field $: d= 7r & «_^r TF -g ©TF where $ TF is the Thomas-Fermi electrical field strength. Its order of magnitude can be estimated, \$. TV i = |*l£l| =^^1 d \yM)\ 1 TF I | dr | r „,. TF b b | djc\~T~] We have introduced here Thomas-Fermi units and expressed the potential in terms of the universal function x(x) . The derivative JL [xj^lj is at x „ l also of the order of ^ty p We find thus that & TF ~ZV. while the Thomas-Fermi radius is of the order Z-\ We get thus or or in usual units, "~ z - z -'-§r,=f- that is, the polar isability is of the order of the cube of the Thomas- Fermi radius of the atom. (This result is completely analogous to the result of the classical theory of electrons for different models of the atom, such as the Thomson model, or the charged sphere model. ) 296 ANSWERS AND SOLUTIONS We note in conclusion that the same quantity for a valence electron is of the order of unity in atomic units, since such an electron moves in a field U(r)tt . Since Z~^$>\, the pola- risability of an atom will thus be determined by the valence elec- trons and not by the Thomas-Fermi electrons. 80.* Let m be the electron mass, $ , r t0 , C , the coordinates of the nucleus, and $ it f\ it ^ (/ = 1. 2, . . ., n) the coordinates of the i-th electron. We introduce centre of mass coordinates, X ~ ~M~\-nm ( similarly for Y, Z) (1) and relative coordinates (i = 1, 2, .... n) x i = U — So ( similarly for y { , Z$, (2) and get then d m d , d d M 2£- <»> dzi M -f nm dX ' dx t ' d£ M-{- nm dX i = i The Hamiltonian of the atom (in the fixed system of coordinates) is of the form H = - /l 2 V' / <?2 <32 ^3 \ , where £/ is the potential energy of the atom which depends only on the relative coordinates (2). If we introduce into expression (4) the new variables (1), (2), and (3), we get for /ytwo terms, one depending only on the centre of mass coordinates, and the other only on the relative coordinates. Separating in the usual way the centre of mass motion, which is of no interest, we get the following SchrSdinger equation in the centre of mass system, In n it 2m Zi * 2M Zd 2d \ dxi dx k "+~ dy t dy k "+" dz t dz k ) ~*~ i = \ i = l ft = l + U(x x xAw{x v .... x n ) = £T (jfi Xn y ATOMS 297 The required effect of the finite mass M (or, in other words, of the motion of the nucleus) is caused by the second term within the braces. Clearly this effect produces a correction (compared to the term s for which M = oo) of relative order -77- • If we take together the terms with / *= k and l^k, and introduce the reduced mass ^ = 7^15-' (6) we get [ n * V a _ ft2 V V (-JL— _i_ * , & \ , \ 2|x Zi* W Li 2d \dx t dx k ~r dy dyk T" dg dz k J ~"~ + U(x t x n ))y( Xl xJ-EV^ x n ). (7) From this equation it follows immediately that the effect of a finite value of M leads first of all to replacing the electron m by the re- duced mass |a and secondly to an additional perturbing term n ^ M ZijLl \dx t dx k ~r dy t dy h ~^ dz { dz k ) ' &> i < k The first of these modifications of the SchrSdinger equation expresses the so-called "elementary" or "normal" effect of the moving nucleus; this effect is also present in the hydrogen atom. It is obvious that the normal effect affects all terms in the same way, and reduces the frequency of all spectral lines of the atom in the ratio H- __ M _,, m m —M-\-m~ l ~J\- The second effect, however, is essentially different for dif- ferent states of the atom . Indeed, first order perturbation theory which is applicable here as 37 <C 1 leads to an energy correction A£=±*J*W< 0) *//'^ 0) rfT, (9) where *i <0) is the wave function of the atom for M = co, and n n rfx =I[ ^ = 11 dXidy^Zi is a volume element of the configuration space of the atom . Substituting expression (8) into equation (9) , interchanging summation and integration, integrating by parts, and using the fact that W -» at infinity we get: 298 ANSWERS AND SOLUTIONS Af = B S S J (8 ' radi ^ * grad * li ' (0)) dz ' < 10 > ,-tO) which clearly depends strongly on the form of W In particular, if the motions of the electrons were completely independent, that is, if the unperturbed wave function of the atom were simply a product of wave functions ^ of separate electrons, n ir (0, = If ?*(*i. y* z i)> (ii) the effect would in general vanish. Indeed, from equations (10) and (11) and the normalisation relation J <picpi rfx i= *» it: follows that: A£ = ^ 2 2 J ** grad ?* dT * J ?* grad ** dT * = °' i<k since both integrals in this expression are equal to zero, (see problem 16, section 5). In actual fact, the electrons in an atom are not independent so that this effect - which in contradistinction to the "normal" effect is called the "specific" effect for a finite mass, and which is characteristic for atoms with «>2 -is, generally speaking, not only different from zero but in a number of cases exceeds appreciably the "normal" effect. The correlation between the electronic motions is caused partly -trivially -by the electrostatic interaction between electrons, partly also, and mainly, by the fact that they are identical particles; this part of the specific effect, which is basically and essentially a quantum phenomenon, can thus be called an "exchange" effect. Indeed, the total wave function of the atom (neglecting spin- orbit interaction, and thus written as a product of coordinate and spin functions) must be antisymmetric in any two electrons. To each value of the total spin S of an electron shell (and thus to each kind of symmetry with respect to the permutation group of the spin function) there corresponds thus a well-defined permutation symmetry of the coordinate function of the shell. Using a rough classical analogy we can characterise this last fact by a prefer- ential motion of the electrons which is either parallel (when the coordinate function is symmetric) or antiparallel (when the co- ordinate function is antisymmetric). Since the centre of mass of ATOMS 299 the atom is at rest it is clear that in the first case the nucleus must move considerably, but only very little in the second case. In other words, in the first case the specific (or rather, exchange) effect will add to the normal effect, while in the second case it will act in the opposite direction. We note that both effects play an essential role for the so-called isotope shifts of spectral lines of very light elements (since the mass Mis different for different isotopes, and the difference will be the more pronounced the lighter the element). A quantitative consideration of the exchange effect of the moving nucleus will be given in the next problem for the case of an atom (or ion) with two electrons. 81. * The neglect of the electrostatic interaction between the electrons is equivalent to choosing for the unperturbed coordinate function of the two -electron system a superposition of products of hydrogen-like functions. Since the total wave function must be anti- symmetric in the two electrons, it follows that the coordinate func- tion must be of the form <F (0) (1; 2) = ^ 1 [c Pl00 (l)c PnZro (2)±<p nZm (l) 9l()0 (2)], (1) where 1 and 2 indicate simply all coordinates of the electrons, while cp 100 =atf and y nlm sav are hydrogen-like electron functions corre- sponding to the given quantum numbers; the first one corresponds to some effective charge Z it and the second one to an effective charge Z (in general Z^Zf ; the indices "i " and "o" correspond to "inner" and "outer" electron, in correspondence with the quali- tative meaning of <p 100 and <? nlm , which describe the ground state and and excited state in an atom); the factor 2 "2 is a normalisation fac- tor. The upper sign in equation (1) corresponds clearly to an anti- symmetric spin state of the two -electrons (para -terms, 5 = 0) and the lower sign to a symmetric spin state' (ortho-therms, S= 1). If we substitute expression (1) into equation (10) of the preceding problem (with /, k = 1 , 2), we get A£ = ^ J j t/*(2)Vu* (1) [«(1) Vv(2)=t«(l)V«(2)ldx 1 dx 1 . The first term gives zero since (see problem 16, section 5) J v*(i). . V'y(/)dT i = Oand the same for a(0). The second term gives A£ = rt^ J* v{l)Vi^{l)dx l J t»*(2)Va(2)dx2= (2) 300 ANSWERS AND SOLUTIONS This is finally the general expression for the required first order exchange correction for a finite nuclear mass. Let us find the selection rule for expression (2). The functions u and v satisfy the following equations (in atomic units): (-i-A-^-) a = £ ia , (3) (_^A-f )<, = £„<,, (4) where *.— &. *.— ?■ < 5 > Multiplying equation (3) by rv*, and the complex conjugate of equa- tion (4) by ru, subtracting the two results and integrating it over the whole of space, we get: I f r(«A^ — v*A«)dx = (Z 1 — Z e ) J j uv*dx + ^ + {E X — Ejjruifdx. (6) Transforming the left hand side of this equation, we get: -i- fr(aAv* — i;*Aa)dx = Y J* rdiv(a Vrf — v* Vu)dx = — — T (" v, °* — v*Vu)dx= v*Vudx where we have used the equations f r div A dx = — J A dx and J grad idx=&idS. (One sees the validity of these easily by multiplying both sides by a constant vector integrating the equation so obtained, and using Gauss 1 theorem.) Equation (6) becomes then (if Zi = Z , so that u and v areeigenfunctionsofthesame Hamiltonian, equation (7) is the same as equation (2) of problem 16, section 5). f v Vu* dx = (Z t — Z e ) [ y u*v dx + (E t — E n ) j rtfv dx. (7) The selection rule for both integrals on the right hand side is the orthogonality relation of the spherical harmonics Y lm (p, 9) which occur in them. It is clear that both integrals vanish, unless ATOMS 301 /„ — /„ = =tl. Since we have assumed that one of the electrons is in a 1 s -state (which is practically always the case), so that / M = we conclude that the left hand side of equation (7), and thus also the exchange correction (2), vanishes unless /„= I, that is, unless the second electron is in a p -state. In other words, the specific (or rather the exchange) effect of a moving nucleus is only present for the P-terms of two-electron atoms or ions (He, Li + Be ++ , ), at any rate, in first order in ^. As far as the sign of the specific effect is concerned, we see from equation (2) that for the para-terms AE > 0, that is, the speci- fic effect has the same sign as the normal effect, while for ortho- terms their signs are opposite. This is in complete agreement with the qualitative conclusions reached at the end of the preceding prob- lem, since in the parastates, both electrons move, roughly speaking, mainly in the same direction, while in the orthostates they are moving mainly in the opposite direction. To find a numerical value for the specific effect, we use equa- tions (2) and (7). From equation (7) it follows that: | j v V«*rfx| a =|(Z f — Z e ) j±. u *vdx+(E l — E n )j xu*vdz^+ + |(Z« — Z e ) J L u * vdlz + {El — E n ) j yu*vd^ + -j-|(Z< — Z e ) J* 1 tfvdx + (£, — £„) j zu'vdz^. (8) The integrals in this equation can easily be evaluated. We express the hydrogen-like wave functions in spherical polars and use atomic instead of Coulomb units (this means changing r(Coul. un. ) -»Zr(at. un. ) /?„i(Coul. un. ) -» z -3/2 /?„i(at. un. )), and we get: «^ioo('O = Kio('-)>w>(0» <P). (9) vzattni m (r) = Rm(r)Y lm (Q, <p), (10) where Y lm is the normalised spherical harmonic and R„i the normalised radial function; in v we shall put /= 1 in agreement with the selection rule, Z e r R^-ft ^ y $±$ 2Z e re'~F (-/t+2, 4, ^)(at. un\ (1 1) R 10 = 2Z'{'e~ Zir (at.un.). (12) 302 ANSWERS AND SOLUTIONS From equations (9) and (10) it follows that the integrals in equa- tion (8) are equal to:- m J yu*vdx=* J* sine cos .pKooK^dQ j R i0 R nl r*dr, o CO fzu'vdx=* fcos%Y w Y lm dQJ R 10 R nl r*dr and so on. o The integration over the angles is elementary, if we use the fact that Y w = -= , sin 8 cos cp = -~y j (Y n + Y u _ x ) and so on, and use the orthonormality of the spherical harmonics. As to the radial integrals they contain, according to equations (11) and (12), exponents, powers, and the confluent hyper geometric function. Looking these integrals up in the standard literature we get finally the following expression (in ordinary units) for the specific effect for P-terms: ** - 3 M^°> {Zin + Zof n + l n C» I) fi2 . where n is the principal quantum number of the p -electron, and Z { and Z Q are the effective nuclear charges for the Is - and np - electrons. These quantities are usually determined by the varia- tional method. For the 2 8 P term of the Li + - ion, for instance, Z { = 2.98, Zo = 2.16. (These values of the effective charges are pretty well the same as those to be expected from elementary con- siderations based on (i) a bare charge of the Li nucleus of 3; (ii) the fact that the dimensions of the outer electron orbit (n = 2) are roughly speaking about 2 2 = 4 times larger than those of the inner electron; (iii) the outer electron has I = 1 and does not therefore penetrate very far inside the inner electron orbit. In other words, the inner electron screens the nucleus from the outer electron so that we should expect Z<^ 3, z th 2.) 8. MOLECULES 1. If we neglect the difference between the centre of mass of the molecule and the centre mass* of the nuclei and take into account the fact that the centre of mass is fixed in the origin of the system of coordinates, the Schr5dinger equation for a diatomic molecule will have the following form: 2m2<[ dx 2^ dy 2-1- dz z) MOLECULES 303 tfl \d / , d\ , 1 d/ . f,d\, 1 s "1, 2Mp*ldpY ap / l~ i ~sI^e^e\ v Sln0 ^8j^~sin2e^J" , " -t-V( Xi , y it Zi ; p, 6, <p)U (...**, y it z it ... p, 6, ?) = £<!>. Here jq, >»i, z t are the coordinates of the i -th electron with respect to a fixed system of reference, the angles 0, 9 determine the position in space of the line which connects the two nuclei, p is the distance between the nuclei, and M the reduced mass of the two nuclei. This equation is difficult to handle in so far as in the potential energy V of the electrostatic interaction, the angles * and 9 occur. In order to put the SchrSdinger equation in a more convenient form we introduce a new system of coordinates, S, % C, which rotates with the nuclei. The C-axis is along the line con- necting the nuclei and the £ -axis lies in the xy-plane. We take the positive S-axis in such a way that the *-, C-, and $-axes form a right handed system . The connection between the old and the new coordinates has the following form: U = — x { sin 9 -j- y>i cos <p, f\i = — jq cos 6 cos 9 — ^cosQ sin 9 + 2* sin 6, {,. = Xi sin 6 cos 9 ~\-yi sin 6 sin 9-f- z i cos 6. If we indicate the difference between differentiation keeping x^ yi, and z t constant and differentiation keeping ^, -r) i( and ^ con- stant by primes, we find: <?Q — d8 "+" Zi V* d^ ^ dZ i) ' i One sees easily that The potential energy in the new system of coordinates will be of the form: V: Z x z^- , y £_ y Z t g2 y z^ P i>k r ™ k = i r *k fc = i r 2ft where 4 = V(6i — y 2 + cu - ^-) 2 -+- (h — c*) 2 304 ANSWERS AND SOLUTIONS is the distance between the j-th and k-th electron in the new coord- inates, while is the distance of the k-ih electron from the first nucleus, and the distance between the k-th electron and the second nucleus. In this way the potential energy expressed in the new coordinates does not depend on the angles 6 and <p. If we take into account all the relations which we have just written down, the SchrSdinger equa- tion will be of the following form: + V<ti. ■»!„ C,; p) — f Uft, f\i, C,; p, 0, cp) = 0. In this equation L z , L^ , and L t are the operators of the components of the orbital angular momentum of the electrons in the ?, v\, C system expressed in units h. 2. We shall denote the spin variable of the i-th electron with respect to a fixed space by s' v and with respect to the moving system by s t . The function <J»(. . . Si . . .) with the spins referring to the 5, y] v C system is related to the function ty(... s^ . . .) with the spins referring to the xyz -system by a linear transformation: *(...*,...) = =. S s(s lt ....,s t , ...,s[, ...,«;, ...)4>(... *;...)= 5-M... < ...). where 5(5 r 5 2 , ..., s., .... «;, .... s'., ...)== MOLECULES 305 and *(h -1)-«4.-t("*>. S (_^_|) = cos | e 4(-T). The SchrOdinger equation which we are looking for will have the form [SHS- 1 — E]<f(...U, ^ *+ s^..; p, 6, <p) = 0, where H is the Hamiltonian given in the solution of the preceding problem . After some simple manipulations we find finally for the SchrBdinger equation, the following expression: -U [i("i)+ e * 6 (i-^)+(A-^) i + +iiFf(a-,- /sine ^- /cos6ir c) 2 ] + -\-V — E]< f (...i i . tli> Zi> Si ...; p, b, 9 ) = Q. Here M^, M v and Af c are now in contra -distinction to the preceding problem, the operators of the components of the angular momentum of the electrons (orbital and spin). 3. We shall assume that the problem with fixed centres is solved, that is that we know the electron terms £ el (p) and the wave functions $ e i. To fix our ideas we consider the a case; let, in the state which corresponds to the wave function <& e i. the momen- tum of the total angular momentum (orbital and spin) along the axis of the molecule be equal to 2. We multiply the SchrSdinger equation £®,dfc. ih. C,;o 4 ; P )/(p)6(e, 9) = = £*el<£i. ■»!<. C i ;o i ; P )/( P )e(6, 9) from the left by <D* e i and integrate over the coordinates with fixed centres and sum over o t . If we note that 306 ANSWERS AND SOLUTIONS we have [ B |(p 2 |)- £el (p)— ^(p)-£ xot +£]/(p) = o, + £ rot B(e, <p)=0. In the last two equations we have introduced the following abbre- viated notation 2Af p 2 " The quantity fi is called the rotational constant. 4. Nitrogen molecule (atoms in state 4 S): 1 S+, 8 2£, 6 2j. 7 2£. Bromide molecule (atoms in state 2 P): 2 1 Z+. 1 2", 'n,. X \; 2"s+, %". 3 n,, 3 U M , X A M . LiH molecule (Li atom in 25 state, H atom in 2 S g state): HBr molecule (Br atom in 2 P M state): 1 S + , *%+, II 1 , m. CN molecule (C atom in *P g state, N atom in 4 S U state): 2 2 + , *Z + , «2+, an, *E, «H. (The number which stands before the term symbols indicates the multiplicity of the terms). 5. The helium atom in its ground state is characterised by the fact that both its electrons are in the lowest state (parahelium). The total eigenfunction of the ground state of the helium atom can be written approximately in the form y=- tya ( ! ) 4»a ( 2 ) {f\+ (°i) ""I- («2> — ~n+ (° 2 ) V- (°i)} • where ty a is a hydrogen function. The hydrogen atom has the eigenfunctions ^(3)%(c 3 ) or <!> 6 (3)t)_(<j s ). If both atoms are at a large distance apart the wave functions of the system can be written as a product ^1'a(04'a(2)^(3){Tl + (o 1 )7 ] _(a 2 )_Tf ]+ (a 2 )7,_(a 1 )}7] + (a 3 ). Taking exchange of electrons into account the wave functions of the MOLECULES 307 system must be anti-symmetric with respect to an interchange of the two electrons . There i s only one anti -sym m etr ic eigenf unction which is also the eigenf unction of our system in zero-th approximation W = ^=^===r {^(1)^(2)^(3)1^(1)^ (2)- — %(2)iu<l)]%<3) + + *a(3)*„(lH»(2) h + (3) ii_(l) — %(l)ij_ (3)]t, + (2)+. + 4'«(2)^(3H 6 (1)[7 j+ (2)t 1 _(3) — t) + (3)7)_(2)]7, + (1)}. (1) In expression (1) = is a normalising factor and 5«= a '/+a0)4'«(2)*»(3)4. o (2)4. a (3)t»(l)rfx 1 rfx 1 rfT s « = /*a0)*«(2)+» (3) * (1)^ (3)^(2) rf^rfx, rfx,« ■ SB /*a(l)4'a(3)*»(2)4'a(2)*a(3)**(l)rft 1 rfx 1 rfx i . If we apply the usual perturbation theory we have * = 2 l fWHVdT:, a where W is the eigenfunction in zero-th approximation and Wthe per- turbation energy, while the summation is over the spin variables. We must take into account that H is different for different parts of * , and indeed for <]> a (l)^ a (2)^ 6 (3) the perturbation energy is equal to A- «.(*.!._-!__ _L + -L + .L\. \R r<x r bi r 6a ' r 18 ' m / '. and for <lv,(l)1»o (3)^(2) it is equal to:- \R r a% r bt r b8 r r u ^r aa J If we take into account the fact that the integrals which differ only in the numbering of the electrons are the same, we have K— A- T=T' (2) where K^e* /VJL-U-L-t-J \ \ My J \R ~ r r 13 " r r 28 r a3 r bl r^J x X<£ (1)^.(2) ^(3)d^dx 2 dx 3 , J \ R r 13 r& r o8 < r 6l r M y ^ X^a(l)^<2)^(3)^(l)^ (3)^(2)rfx 1 dx 2 rfx 3 . 308 ANSWERS AND SOLUTIONS The integrals K, A, S are in general of the same kind as the cor- responding integrals in the hydrogen molecule problem. An evalu^ ation of the integrals shows that equation (2) corresponds to a re- pulsion. This is correct not only for helium but also for all inert gases. 6. If we split off the motion of the centre of mass we find for the wave function of the relative motion of the nuclei the following equation A* + !( £+ 2D(i-J# = 0. If we separate the variables in spherical coordinates and put It^Zf-YsMi*, cp), we find for x the differential equation g+H+ ^ + T +1) ]x = o. If we make the substitution x(p) = P s<?-Xp "(p). where s = -^-±- + y T 2 + f*' + y) » the last equation goes over into the hypergeo- metric equation pa" -+- (2s — 2Xp) u' + (— 2sX + 2 T 2 ) u = 0. The solution of that equation which is finite for p = is of the follow- ing form u=r-cF(s — t- t 2s, 2XpV In the states of the discrete spectrum the wave functions x must tend to zero as r-*oo. This means that the expression for u must become a polynomial, t x Y" „ where v is a non-negative integer. From this condition we get the energy levels E q k = Jfi y 4 2p& [• + T + /> + (*+±)']*' MOLECULES 309 The dimensionless parameter f- is proportional to the reduced mass of the nuclei jx, so that y 2 ^> 1. If v and K are not very large, the expression for E vK is of the following form E r ir = -D-f- /*«>„(*; + !) + + where 2fx«H A ^2; 2^r + 2j 2J^H ' i/~2D The dissociation energy is approximately equal to The second and third term in the expression for E vK gives us the vibrational and rotational energy. The fourth term takes into account the anharmonicity of the vibrations and, finally, the fifth term gives the correction to the energy taking into account the in- teraction between the rotation and the vibration of the nuclei. Since D is of the order of unity in atomic units (e= m = h = 1), it follows from the expression which we have obtained that where m is the mass of an electron. It is thus clear that the difference in energy between two quan- tum states with different electronic motion (a quantity of the order of D ) is large compared to the difference in energy of different vibra- tional states which in turn is large compared to the difference be- tween rotational states. 7. We can find the minimum of the effective potential ir = _2D(±- i) + £, where ^=^^+0 by putting its derivative equal to zero W' = — 2d(~ -t-h-4-) — ^ = 0, V Po ?V Po 310 ANSWERS AND SOLUTIONS from which we get i ^ A * We shall now expand the effective potential near the equilibrium position which we have found U7(p)^-2D(-L-4-) + ^ + ^4^( P -p )2, vp o Po 7 P5 Po and retaining only terms up to the order A 2 , we get ^(p)«-D + A 2 + (D ~ 3^ 2 ) (P — Po) 2 - We can substitute this expression into the equation for x $1 + 1? [E + D — A* — (D-3A*)(p-tf] X =0. We find thus for the energy levels In the same approximation here Finally, we have for the energy E vK E vK = — D + Ao> (v -f 1).+ ' 2(jL«a 2 p. 2 a4w By virtue of the fact we have not taken into account in these calcu- lations the effect of anharmonicity, we have not got here a term — ^uPY^ 2") w hich we found in the preceding problem. 8. In the infra-red band we are dealing with transitions in the electronic ground state where vibrational and rotational quantum numbers are changing. For the frequency of the transition between two states v', J'-+i/', J" we have From the selection rule for / it follows that J" = /±1. We get thus the following frequencies MOLECULES 311 and -m^-o-^+i, (;: y ;tt..) We note that these two sets of frequencies are in molecular spec- troscopy called the P and R branches. From the expressions which we have obtained it is clear that the difference in frequency of two successive lines for given values of v' and 1/' is equal in cm -1 . Aw h The moment of inertia of the HC1 35 molecule is equal to If we use the value of the reduced mass, fi = Af H ^A = 0.972^ H =1.61 . l(T 24 g we find for the distance between the nuclei in HC1 , a = y — = 1.29 - 10 _8 cm. The equilibrium distances a in the molecules DC1 and HC1 are the same since the form of the potential curves is determined by the electronic states. It follows thus that Av DCl I^HCl . A ir . ■,„_-! Av D ci= 10.7 cm . Av HCl M-DCl 9. The distance between the first two rotational levels is given by A %t= 2^/= 41.5cm- We find thus Av . ^-rot^ 0.0104. vib 312 ANSWERS AND SOLUTIONS 10. The dissociation energy of the D 2 molecule is equal to 4.54 eV. 11. If we go over to a new variable $ = r ~ a . , we can write o, the equation for the radial function y as follows; If we put z = ae- 2 fc, we find *''^x'+U-4+^i)*=o, where 5 = -|/" ^< D - £ ) , + s ii_l^ ff2 _ 2tufiD If we make the substitution y = e' 1 z 8 u (z) this equation will go over into the hypergeom etric equation zu" -f- (2s -+- 1 — 0) u' -f- rcu = 0, the solution of which is the confluent hypergeom etric function u = F( — v, 2s -\- 1, z). This function satisfies the condition that -/ tends to zero as r-*-|-oo for positive values of s (discrete spectrum). As r -> — 00 the wave function must tend to zero. In order that that condition is satisfied it is necessary that F reduces to a polynomial, that is, that: v is a non-negative integer. This con- dition determines the energy spectrum £ - J *(«+i)-w(«+r) ! ' wte — 4 ?/^- In this way the difference between vibrational levels decreases with increasing quantum number v. The dissociation energy is equal to fi(0 . fi 2 0) 2 E n = D- 16 D 13. As parameters to characterise the rotation, we shall use the Euler angles (6, <j», cp). In this case the coordinates of a point x, y, z in the fixed system are connected with the coordinates $, f\, C in the moving system by the following equations x = $ (cos ty cos cp — sin <]> sin cp cos 9) — — i] (cos <]i sin cp -j- sin 9 cos 9 cos 9)-f C sin 9 sin 6, (1) V = % (sin 9 cos cp -f- cos 9 sin 9 cos 9) -j- MOLECULES 313 + "*)( — sint{>sin<p-|-cos^cos<pcos6) — £cos<j;sin9, z =?sin<psin6-}-7)Cos<psin6-(-Ccose. To find the form of the operators \, J v j\, we use the fact that the operator 7 e is 7 e = — /^-, where a is the angle in the plane per- pendicular to the £-axis. Since under a rotation of the 5, % c - system with respect to, for instance, the £ -axis over an infinitesi- mal angle da, the values of the angles change, we can write for y 5 , in units h, 5 \da <?8 "^da d<?*~ da di?) For an infinitesimal rotation around the % -axis over an angle da. we have 6 = 6', f\ = ff — C da, C = ij / (/a-|-C / (2) and z = %' sin («p -f d<p) sin (0 -+- dO) + + if cos (<p + dcp) sin (0 + d9) + t/ cos (6 -f dfl). ( 3 ) On the other hand if we substitute expression (2) into equation (1) we get z = V sin cp sin 6 -j- if (cos cp sin 6 -f- cos 6 da) -f- + C (cos 6 — cos (p sin 6 da). (4) Comparing equations (3) and (4) we find ^-=coscp, ^ = -sincpctg0. Similarly we have Q = i!"_l da sin 8 We find finally for yj the expression •4— '(™»w--"T«*^+£f^)- 314 ANSWERS AND SOLUTIONS In the same way we find expressions for the two other operators ? . ( . d , b d , cos cp d \ 16. Ej=£jiJ+l). Each energy level is 27+ 1-fold degenerate with respect to the direction of the angular momentum referring to a fixed system of reference and equally degenerate with respect to the direction of the angular momentum referring to the body itself. 17. Since H = ± / 2 +- 1 (^ — -[-) J*, we have E = -^ • . ^-hO-h^-f^- — ^)^ 2 ' y c = * l*l</« In this case the energy levels are 2(2J+ l)-fold degenerate. The degeneracy as far as the direction of the angular momentum in the fixed system of ref- erence is concerned is 2J + 1-fold as before. 1ft fig | 1 d ( . „du\ , 1 ( d 2 u , d*u\ i0 * ~" 2J \HnTd8 V Sm M J "^ sin2 6 ^ d=p "+" d^j 1 fi 2 / 1 1 \ d 2 u _ F sin* 8 dp? I 2 \C .4/ dtpa ~ cos 8 d 2 u 19. Since > commutes with 7 C =— /-^- and i g = — /^ we can write the eigenfunction in the form where Mj and k are the components of the angular momentum with respect to the fixed z-axis and the moving £-axis respectively. Since <|> and cp enter symmetrically inequation (14) and |&|<7, we have | Mj |< J . Let us consider the operators 4+4 -—"-•'(a— <«*•£ + «£,£). MOLECULES 315 One can easily show that A (A — it) **JMj = ( k + l ) ( y e — "J ^kJMj, i.e. the expression O i —U 1 )$ k jMrj is the eigenfunction correspond- ing to the value (k-\~\) of the operator i c . If we put k = J, we have {j\ — ij\)$JJ Mj ==Q. The last relation can be put in the form (£+' d *'£-in£) e '"'<»' uw « 4 *- - We find thus an ordinary differential equation of the first order to determine Qjjmj d&jjM T Mj — J cos 8 ~rf6 1 skTO *J™j = 0. The general solution of this equation is of the form (sin 6/ VJJMj = c Mj K) or J-Mj J+Mj e(8) = c(l— cosO) 2 (1-|- cos 8) 2 (3) Since the function 6 must be finite, we have |Af J |<i. To determine the function Qhj-mj, we consider the action of the operator (ig + ii^) on it. Since A (A + VJ ®* JM J = (* — * ) (-4 + '•£») e * J*.r we have (^ + '/) ®kJMj = O-kftk-UMj- (4) Substituting into equation (4) the explicit form of the operator (jl+t/,) from expression (1) we are led to the equation which, if we introduce a new variable, x = cos 8 becomes of the (5) 316 ANSWERS AND SOLUTIONS form dP,. nt (x) Mj — kx Y i _ X 2 L_£ |_ />*./*,(*) = — lafk-VMj (v), r/jr y 1 — a- where ^Jji/y (a-) = Q k ,/ Mj (arc cos x). If we put P/tJAIj — (1 A') 2 (1 + -V) - 'U W j|/ >/ , we find a simple relation to determine v UMj dv hJM T —j^ — = la-kVk-UMj- (6) The function Pjjvj which we found earlier (see equation (3)) can be written in the form of expression (5) ( J -™J) (J + X.T) PjJMj(x) = (l —X) -' (1 -f x) 2 VjJUj. where we understand by v JJMl the expression vjjmj (x) - c ( " 1 - x) J - Mj (i + x) J+Mj . (7) From equation (7) and the recurrence relation (6) it follows that V kJMj=C—-j-^ Hence <>" " {( i _x) J - M '(H-*) J+iU '} Je-Mj k+Mj == ce ik,f e iMj * (\ — cos 6) 2 (i+ C osO) 2 X \«7*-*: x(^STe) " {(i-cose/-^(i4-cos6/ + ^} ; for Mj = these generalised spherical functions go over, as we should have expected, into the ordinary spherical functions and are the wave functions of the rotator *wo(e, cp) = ce^ -±- **'* (sink 6). sirr 8 (rf cos or MOLECULES 317 20. #*k+2 ~Ik+2 k = — #*+2 k — 21. For an asymmetric top the degeneracy referring to the direction of the angular momentum with respect to a fixed system of reference remains the same. The degeneracy with respect to the quantum number k disappears completely so that a given value of / corresponds to 27+ 1 different levels. In the case / = 1 the energy levels are determined from the solution of a secular equation of the form H n -E #10 "l.-! #10 #oo — E #0-1 #-n #-10 #-i,_i-£ H u — E "l.-l #oo — E "-i.t #-!,-!-£ = 0. Since #_i f _i = # u , we have and thus (tfoo — E) (#11 + E 2 — 2H n E — Hi _i) = 0, *.-!M- *-?£+;}-). *.-!(*+*)■ 22. E m Ex $ fc=0, »» 1 -^ — e*<\/ sin 8 4ji 1/| cos 9 — 1 \$. *-*+ sin 6 4it 318 ANSWERS AND SOLUTIONS £2 + ®k=-l,tn) 1 — 1 — Y Tc-e*^ (cos <p -f- / cos sin <p) — 1/ -£■ cos <p sin 8 1 AT — 1/ -rz.e-*^ (cos <? — /cos 8 sin «p) Ez — *ft=-l,m) 1 — 1/ t^ ^ (cos 8 cos <p + f sin 9) 1 AT — 1/ -5- sin 8 sin <p ft r — 1 -1/ Tt e— 1 * ( — cos 8 cos <p + * sin f) 7t r lb 24. The splitting of the terms is determined by the spin-spin interaction. To determine this splitting it is necessary to average the operator of the spin-spin interaction, a(S») 2 , over a rotational state. For a given value of K the quantum number / takes on the values J=K-\-l, K, K—\. The matrix elements of (»£) which are different from zero have the form y nS >K j^R — WSh-u-K^V 2AT+1 ' \ n!i >K J=K — { < n *>K+lJ^K—V 2/C+l * \- n *>K+i j=k+i — y n *)s: j=k+i — V 2/C+3 ' /„C^ J=ff- 1 /-.cx^-l J^-fiT-l -.A /C— 1 l"* JjT-1 ,7=fi;-.i — K*a)g j=K-l — V 2K—1 ' Using these last relations we find for the splitting of the components of the triplet A£j = k +i = 2K + 3 a ' ^ E <r= K ~ a ' ^ Ej = K ~ l = 2/(— 1 * ' 25. We find first of all the off-diagonal elements of the op- erator w. t \w 'wA2W MOLECULES 319 One sees easily that the only matrix elements which are dif- ferent from zero correspond to Q' = Q ziz 1 , J = f. Since {^as ± i == + * w"i* ± t , we have . nA2« x{iK\ ' nA.Sl ± Iv To evaluate the matrix elements we note that if we put in the operator (Jz+U r ) (see problem 13, section 8) cp = and — ij- -► M v we have If we evaluate this last equation we have {^Jr+iinls+f 2 * 1 )'* 6 ! hSJ A2 ± U = ±/-K(yzF2)(7ri:Q-hl). In the case of small vibrations near the equilibrium position the matrix element {i „ . nA2» P 2 n W ± lv can be put approximately equal to 1 I mm lJtA.2 TF l^/nASt 1' where p is the equilibrium distance between the nuclei. Since ii, + *,) SS t . = (5,C ± , = =*= T /S(S+ D-i)(^l), we have finally: 320 ANSWERS AND SOLUTIONS \w ni.2 V J I »A2 ± IvJ = B VS(S+l) — I(Sz£l) j/"7(7-h 1)— (A+I)(A-f lirl. Here B = h is the value of the rotational constant in the equi- 2jWPq librium state which corresponds to some p = p . In the general case the doublet splitting may be of the same order of magnitude as the matrix elements we have evaluated. If we consider, therefore, the displaced levels of the doublet term we can apply perturbation theory in a somewhat different form. Instead of the functions YrtA A + i/vvJ t 't'nAA - \!tvJ we use as zeroth approximation a linear combination of them If we substitute this expression in the perturbed equation and use standard methods we find the secular equation c(0) p ,T„, nAA + l k vJ £ n A A + i/tt) J c w nL A - %vJ = 0. nAA-i/stJ" p(0) P W«AA + i/ivJ £ wAA - i/,« J £ From the solution of the secular equation it follows that f = l£ (0 >±i- 1 /"A^ > + 4^{(y + ^) 2 _A 2 }, (1) where ^ — c nAA + i/ iV J -f- £ wAA - i w« AF (0) — F (0 K p(0) In the a case when the multiplet splitting is large compared to the rotation, it follows approximately from equation (1) that C 1 — c wA a + i/j„J" -) — j^ , ^2 — £wAA- Vs»J £!{(/+ 4)' -A'} In the b case we get from equation (1) fl , 2 =i-^>-B {(y + |) 2 -A 2 }± A£ (0) 8S {(/+y) 2 -Aa) MOLECULES 321 26. K = 0, 2, 4,..., if the total spin S = 2, or 5 = 0; /C=l,3, 5, . . . , if the total spin S = 1 . 27. The magnetic moment of the molecule is equal to eh 2^(A-f2S)«, where n is a unit vector in the direction of the molecu- lar axis. To determine the splitting of the energy it is necessary to av- erage the quantity - 2 4 c (A + 2E)»|C over a rotational state, that is to determine the matrix element Since / is a vector quantity which is conserved, it is clear that the matrix elements of the vector n must be proportional to the matrix elements of J, that is, 3 3 If we consider n as an operator we have » = const J. To determine the constant we multiply the last expression to the left and to the right by J. Since the eigenvalues of J 2 are equal to y(y-|_i), and Jn is equal to Q, we have In this way the operator of the perturbation energy is equal to If we evaluate the diagonal matrix elements we get for the energy splitting the following expression ^ = -& + 2S >7<JTT7^- 28. The perturbation operator in the given case is, as one can easily verify, of the following form : It follows from this that the Zeeman splitting is equal to 322 ANSWERS AND SOLUTIONS v i A2 >/(/+l)-S(5-H)4-/C(.<C+l) , J(J+l)+S(S+l) -K(K+D ) *\ 2/c(/c+i)/(y+i) "t J{J+\) 1" 29. The energy of the Zeeman splitting is equal to 30. Since the interaction energy of the magnetic moment with the external magnetic field is of the same order of magnitude as the spin-dipole interaction it is necessary to consider the two simul- taneously. The perturbation operator has the form V = AnS — |x Ai*S£ — 2^% . As zero-th approximation wave functions we use the wave functions of the states in which the angular momentum K and the components of K and S along the direction of the magnetic field have well-de- fined values. The «-axis is taken along the magnetic field. Since the component of the total angular momentum along the direction of the magnetic field is conserved we can, for the case of a doublet term, apply the perturbation theory for two -fold degeneracy. If we evaluate the matrix elements of the perturbation operator we have v *k. -■;. = — M * 2KTK+T) A — M * KTK+T) V-" 36 + *<&*• A 2 v M K _!/, — 2 ' * in ylM K — 1 A i i • \ M * = ^A{n x ^-in y ) MK _ l , Substituting those expressions into the secular equation and solv- ing it we get MOLECULES 323 We consider two limiting cases. K H&e~^>A, we get for f£> 2 the expression which coincides with the equation obtained in the preceding problem, For A^^sv we get (i) _ AA l 1 w A 2 m+i)~~{ MK —j) L„ , .!?.. , i v -i — 1 -t\h&. >\ {K +l)(K+\) + K+\ \ The second terms of these equations, that is the terms which are linear in 36, agree with the corresponding expressions which we ob- tain if we substitute in the equation found in problem 28, Section 8 7= K± V 2 , 5 = Va and.yW, = M K — 7 2 . 31. Because of the axial symmetry the dipole moment of the molecule is directed along the line connecting the nuclei, that is P=pn. If we proceed as in the solution of problem 27, Section 8, we find A ^ = -^7(7TTj-% 32. 3 i *K{K+\)-J(J+\) * 33. The first -order correction to the energy is equal to zero. It is well-known that the second-order correction to the energy in a case where there is degeneracy can be found from the condition that the simultaneous linear equations c n C n? = 2j CkJ 2^ 1/nP yjna * n< v nj 3 ma n c m have a solution. In our case we find from this relation h* \ (2/ + D(2/-i)z (2< + a)0tf + i)(/ + i) )- In this way the energy of a rigid dipole is equal to 324 ANSWERS AND SOLUTIONS This result is paradoxical on several accounts. Indeed, accord- ing to the last equation, the energy of a rigid dipole is not propor- tional to t, but to i 2 . We can thus formally assign to a rigid dipole " polar isability". Let us consider the case when / = 1. The level corresponding to the value m = 0, has in an electric field a larger energy than out- side it so that the corresponding molecule moves in such a way as if it had a negative polarisability. It behaves as a diamagnetic solid in a magnetic field. A molecule in the state m = ±l behaves "normally". The degeneracy in the electrical field is lifted only partially , since the energy depends only on the absolute value of the component of the angular momentum. m = +l We note that 2 £?Jt = 0. m = — I 34. For large values of R we can neglect exchange effects, that is, assume that the first electron is near the nucleus a and the second electron near the nucleus b (see fig. 42). The interaction between the two atoms which is of the form v = ±— L__L + i- must be considered as a small perturbation. In the first approximation the interaction energy of the two atoms is equal to the diagonal matrix element of V, that is Fig. 42 where / <to ( r to) % (r 2b ) V% (r 10 ) % (r a6 ) dx t dt 2 , In the S -state the diagonal matrix elements, that is, the average values of the dipole, quadrupole, . . . moments, are equal to zero so that for an evaluation of the interaction energy it is necessary to go over to second-order perturbation theory. In the perturbation operator (1) we restrict ourselves to the dipole-dipole interaction as it decreases most slowly with distance. To obtain the operator of the dipole-dipole interaction we expand the potential V in decreasing powers to R. An expansion in spher- ical harmonics gives MOLECULES 325 1 1 vi f x ^ = |/?P-r la | = 2^x(cosft) = r bi X = or' /?» ^ 2/?3 r i2 i /?p H- r 62 — r al | = _L i (fat — r b2 ,p) . 3(r al — r b2 , p)2— (r a1 — r 62 ) 2 , _L_I + ( r 6 2 P) , 3(r 62 p) 2 -rg 2 ( /? ■ /?a -r 2^3 Substituting this expansion into V we find an expression for the dipole-dipole interaction v — 2 *i*2— -*i-*2— yi-ya /ov y?3 » (^) where the z-axis is taken along the line connecting the nuclei. We have already shown that the average value (2) over the unperturbed eigenfunction <1» = <fo (r al ) <|» (r 62 ) is equal to zero. The non -diagonal elements of (2) which correspond to a transfer from the ground state to excited states can be written in the form 00 /J3 • From the selection rules we see that the matrix elements z^ jc 0n , y Qn are different from zero only for transitions from the ground state to the states <}>„(/•) cos 8, ^ ( r ) sine cos <p, where these three matrix elements are equal to each other. The interaction energy in second approximation is equal to F < 2 >_y <^") 2 _ 1 y ^lnZln + xlmAn + ylmyln Lk 2E ~E m -E n - *e 2d 2E -E m -E n or fi 2 2 /?« 2d2E -E m -E n (3) 326 ANSWERS AND SOLUTIONS Since E < E m and E < E n , E® is negative, and we find thus that two atoms in a non -excited state which are at a large distance apart, attract each other with a force which is inversely proportion- al to the sixth power of the distance. To evaluate the sum in (3) approximately, we note that the difference in energy between differ- ent upper levels is small compared to the difference of energy be- tween the ground state and the upper levels. We can, therefore, write expression (3) approximately in the form /?« £ Zom 2u E -E n ' w» n From the theory of the quadratic Stark effect it follows that _2 j V _ °" D = — — q, where a is the polarisation of the atom. For the hydrogen atom ground state a = 4.5 atomic units. To evaluate the sum 2*o M we use the matrix sum rule JM±0 (AB) nk = ZJ A nm^mk m or m Putting in the last relation A = B = z, n = k = 0, we get (2 2 )<X) — 2j z Om z mO == 2j z 0m m or m jk »» Since in an 5 -state * 00 == and (* 2 ) 00 ^(r^, because of sym- metry, we get We find thus finally for V(r) the expression V(R) = -^. MOLECULES 327 To ascertain how the interaction forces between two neutral spherical symmetric hydrogen atoms arise, we consider the wave function of the system. For the wave function of the system we find in first approximation *t = % ( r ai) ^0 (r b2 ) [l + 2£T£3 (x t x 2 -h y t y 2 — 2z t z 2 )J . The probability density «(1, 2), has, if we neglect terms multiplied by R~*, the form w (Tav r b2 ) =*= w (r fll ) w (r M ) [i -f- -^ (x t x 2 -hy^y,— 2z x z t ) J . If there is no interaction between the atoms the probability density is simply equal to the product of u>(l) and w (2), that is, there is in that case no correlation whatever between the positions of the electrons. In the case of an interaction the position of the first electron is not independent of the position of the second one. The electrons occupy statistically more often positions where the mutual potential energy has the least possible value. The interaction force can thus in first approximation be ex- plained not by the deformation of the electron clouds but by the cor- relation between the electron positions. 35. Let us prove the additivity for a system consisting of three atoms. From our calculation it will be clear that it can be applied to an arbitrary number of atoms. We write the interaction energy in the following form V = V(1, 2)+V(2. 3) + V(3, 1), where we have denoted by 1, 2 and 3 all the coordinates of the first, second and third atom. We consider the interaction of atoms which are at large distances apart. In that case exchange forces play no role whatever. If we ignore exchange forces we can put the wave function of three atoms in zero-th approximation in the form ^ = ^i(l)^fc(2)^(3), where /, k, I indicate the quantum states of the atoms a, b, c. The functions ^ ai (l), belonging to different values of f, are ortho- gonal. The same can be said also about the functions ^ bfc (2) and *«i(3). The perturbation energy in second approximation is of the form 328 ANSWERS AND SOLUTIONS vT,\ 2 yooo , Vi' J Vug] "ooo-T^ E a0 -\-E bQ + E c ,-E ai -E hk -E cl % U) The prime on the summation sign indicates that /, k, I cannot be simultaneously equal to zero. The first term is the classical mul- tipole interaction. In our case it is equal to zero. In expression (1) all terms where at the same time i =£ 0, k =£ 0, / =£ 0, disappear because of the orthogonality of the functions. Three partial sums with t = k = 0, 1^=0; t = i = Q, fc =£ 0; k = l = = 0, j'#0 refer to the polarisation interaction of the l-th, fc-th, and *-th atoms in the resulting field of the two remaining atoms. In the case where the distribution of changes in the atoms has spherical symmetry these sums are also equal to zero. It is nec- essary to remember that these sums cannot be obtained by taking additively into account the interaction energy of each pair of atoms. We must finally consider those terms where two indices are (lif- erent from zero. Thus, if we make again the same assumption about the charge distribution in the atoms the interaction energy can be expanded in three partial sums vTJ 2 . *v \vZ\ 2 JU Eao + Eto-Eai-EM^ -J Eao + Eto — Eai — E b ]c *U £ 6o +£ c0 — E blt — E cX + 2 (vIS?) 2 Eao ~r ^co E a { E c i (2) Because of the orthonormality of the eigenfunctions of the atom we get for the matrix elements V^=/faoO)^o(2)^(3). {V(\, 2) + + V(2, 3) + V(3, 1)} ^(1)^(2)^(3)^^ = ■=/4Jo(l)4io(2)V(l, 2)^ ai (l)^(2)rfT lC fx 2 ={K(l, 2)}5J. Expression (2) consists thus of three terms, each of which is the dispersion interaction of a p?ir of atoms. One sees easily that these considerations can be extended to an arbitrary number of atoms. In the case when the distance between the atoms is not very MOLECULES 329 large, one must take into account the transition of electrons from one atom to another, that is, exchange forces. 36.* The basic physical fact in the quantum mechanics of mole - M cules is the very large value of the ration — , where M and m are respectively the nuclear and the electronic masses. Indeed, the presence of this large dimensionless parameter, of the order of magnitude of 1000 to 10, 000, causes a considerable difference be- tween the orders of magnitude of the quantities mentioned in the present problem . Let a be of the order of magnitude of the linear dimension of a diatomic molecule. The distance between the nuclei will clearly be of the same order of magnitude. Indeed, it cannot be larger than a because of the meaning of this quantity but it can neither be con- siderably smaller than a because of the mutual electrostatic repul- sion of the nuclei. (We can easily arrive similarly to the conclu- sion that o must be of the order of magnitude of the linear dimensions of an atom. This fact about the order of magnitude of a does not play any role in the estimates in which we are interested. ) 1. We shall first of all estimate the order of magnitude of the energy of the valence electrons and at the same time the order of magnitude of the intervals between electronic levels. Since the valence electrons move, in contra-distinction to the electrons of the closed inner shells at each of the nuclei, in a region of space of linear dimensions — a, so that the uncertainty in momentum Ap will be of the order of magnitude hfa, the zero point energy of the electron, or the difference in energy of successive electronic levels, will be of the order of magnitude E e j: Cel m H^' (1) We shall now consider the vibrations of the nuclei in the mole- cule. We can use as a model, at any rate for the ground state and the lower excited levels the motion of a harmonic oscillator of mass of the order of M (or rather, with the reduced mass of the nuclei) and a stiffness coefficient K. We can estimate the latter from the fact that a change of the distance between the nuclei of order a must correspond to a change of order unity in the electronic wave function, that is, it must be connected to a change in energy of the order of magnitude Ka*~E eh 330 ANSWERS AND SOLUTIONS so that we get in the usual way for the frequency of the vibrations of the nuclei in the molecule <o~ j/ -^ and using equation (1) we get for the intervals between vibrational levels Finally the rotational levels of the molecule can clearly be treated as the levels of a rotator with a moment of inertia I~Ma 2 , so that we get for the intervals between rotational levels ^rot^ / ~ Ma* ~ M e1 ' l ' It is clear from equations (1), (2), and (3), that the quantities £ e j, ^vitand E I0t form a geometric progression with the factor (j^) '~ 10~ : 2. Let b be the amplitude of the zero point vibrations of the nuclei in the molecule. It will be of the order of magnitude of y _L (we can obtain this estimate in different ways, for instance, by equating the order of magnitude of the oscillator energy, (~h*) and the potential energy for a displacement over a distance h (^M<o2* a ) which from the expression (2) for a> gives us 'Km) a ' (4) The ratio of the vibrational amplitude b to the equilibrium distance a of the nuclei in the molecule is thus of the order of mag- nitude of (4r)' *<C l - We can consider this quantity to be a small expansion parameter in the theory of molecules. According to the results obtained earlier £ el , E vih and E Tot axe quantities of respec- tively the zero-th, second, and fourth order in this smallparameter. 3. The periods of the electronic motion and of the nuclear vibrations in the molecule are of the following order of magnitude — 1 ft mcP /c\ T e \ — b — j— , \°) Cl «>el E el S The corresponding characteristic velocities are clearly equal to ei / el ^ m J ™& ; vib' vib SCATTERING 331 (or ~/^)~-^. (8) From equations (5) to (8) it follows that The last inequality means that the nuclei are moving slowly compared to the electrons in the molecule. This fact makes it possible to consider the nuclear motion in the adiabatic approxi- mation. 9. SCATTERING. 1. The potential energy of the particles is U(r) = — U (r<a), U\r)==0 (r>a). It is necessary to find the phase shifts, that is, the asymptotic form of the radial functions satisfying the equations r>a x? + [* 2 _'J(±>)] X( = o, *-..%*. and with the boundary condition xi (°) = °- When the de Broglie wave length is considerably larger than the dimensions of the well, the main contribution to the scattering arises from the S -wave. The solution y , which satisfies the boundary condition is of the form Xo = A sin k'r (r < a), to = sin i kr -h s o) ( r > «)• The phase 8 and the coefficient A follow from the condition that both the wave function and its derivative are continuous at r — a. We get in this way 332 ANSWERS AND SOLUTIONS & = arctg(^rtg*'a) — ka. The partial cross-section for / = is thus o = -£ sinU = % sin* [arctg (£ tg k'a) - *a] . (1) For small velocities of the incident particles (£-»>0) S will be proportional to fc Because of the factor 4r the cross-section a w* 11 ^ « 4ita 2 (-^^ — 1 ) 2 (for small A) (3) We consider the cross-section o as a function of the well depth, which determines k . If the well is shallow (M<C 1). we have ky i6* awy *o==W-_- = _ — __. We note that we get from perturbation theory /<»>=-i-lH><*=T< and thus The cross-section increases with increasing u and diverges for k a = -J . The condition fe a = y is the condition for the appearance of the first level in the well. If we deepen the well even further, the cross -section starts to decrease again and tends to zero for tgk a = k a. When U is further increased the cross-section continues to oscillate between and oo, becoming infinite whenever a new level appears in the well. The sharp oscil- lation of the cross-section for the scattering of slow particles explains why the cross-section for the scattering of slow electrons by an atom can differ appreciably from the geometric cross-section. SCATTERING 333 We note that if k a is near to an integer times £tt, we must alter equations (2) and (3). Indeed, in that case igk'a is a large number and we cannot use the expansion which leads from equation (1) to equation (2). In that case we can still neglect the term with *fl<l in the square brackets of equation (1). We have thus a = arctg[Atgfc' fl | and we get for the cross-section <j where *3-|-#» k' 1 This equation for resonance scattering gives the dependence of the cross-section on k for small values of k, for the case where the potential of the well is such that a small change in its depth or width will make a discrete level appear or disappear. 2. =w(i^-i)\ „W*=« If u -+oo a = 4rca 2 , that is, it is four times as large as the elastic scattering cross-section for an impenetrable sphere in classical mechanics. 3. oo , 6 cos d 1=0 2('-f l)sinS z sin8 z+1 cos(8, +1 — 8 z )-f 1=0 CO 5 3cos»»-l y f /(/+!) (2f +!) _,_». , 2di (2/-l)(2/ + 3) Sin 6 * + * 2 2 Ai\ (21— I) (2/ + 3) or "+" 3(/ t/+ ( 3 +2> sin *i sin V* c «s (8, +2 — 8,) } + . . . f A °° J rfo==J SS( 2/ + 1)sin2 ^ 334 ANSWERS AND SOLUTIONS- * OO f cos i\ do = ^ J] (/ + 1) sin 8j sin o, +1 cos (8 m — 8 Z ), U 1 = f 3 cos2&-l _ 4* y /(/ + ■!) (2/ + /) ■ 2 , , *=o CO + JgL J itjim sin 8, sin 6, +2 cos (o 1+2 - 5,) . 7=0 4. The radial function satisfies the equation and the boundary conditions Xj(°) = ° and X = finite as r -► oo. The solution satisfying these conditions is where From the asymptotic behaviour of J x (kr) we get the phase shifts, ^-fH-IHf{/(<+i) J +^-(<+T)}- The independence of S z on A: means that we get for the scattering amplitude /(». *) = |/oOU where / (fl) is independent of the energy of the scattered particles, The scattering cross-section, is inversely proportional to the energy and is characterised by a universal angular distribution. Since the sum oo /(») s =wS< 2/+1)/,|(co80)leW8, ~" 11, 7=0 SCATTERING 335 which determines the scattering amplitude diverges for &-*o » it is clear that large values of I are essential for the evaluation of ^(f>) at small values of . For large values of I we have _a.~ «M <:u ( 1) 1 (2/ + 1) IP so that np.A 1 l=o 2sin l" -^<ri expression (1) for 8 ? is valid for all / and thus for all values of JK f )~ kh % b • 2sin T d3== W Ct S2 d!} - 5. The scattering amplitude is in the Born approximation given by the equation W» — -S&r Ji*i/M* — 3£. where q = k' — k, g = 2ks\n-^. Hence ^Bon,= l/( ft )N2=^ctg4d». In classical mechanics we have the following connection between the angle of scattering and the impact parameter p pupdr rc — ft f J r*y r 2 i x(E-U)-(^) 336 ANSWERS AND SOLUTIONS where r is the zero of the expression tinder the square root sign. If we integrate, we get 2 _ A 1 (tc — 0)3 P ~~ E ft 2k — 6 ' and thus If d3 = - 2*p -^ «fl> = — a 2(27C _ 0)a rf»- we can apply the Born approximation for all angles (see preceding problem). In the opposite limiting case, where -^^>1 the classical result holds for not too small angles, while for smaller angles, the Born approximation result is valid. 6. We have the following equation for the radial function, *+t£U+<V"")x = o. Using the notation k 2 = --^j- , x 2 = -^ and the independent variable r Z = e~** r we must solve the equation its solution is a Bessel function of imaginary order x=J± 20H (2ax£). The function x must vanish at r = 0, that is, at 5 = 1, so that we get, apart from a normalising constant, 7. = J-taki (2ax) J 2aki (2ax£) — J 2a u (2flx) ^_ 2aft i (2ax&). (1) The asymptotic form of x as r->oo (S-+0) is SCATTERING 337 e 2aki\n.ax g -2akiliia* * = J-.au (2ax) r(2a , /+1) e-*^j 2ak . (2fl x) r( _ 2fl , f+1) ««*. The coefficients of e~ ikr and e ifcr can be considered to be functions of a complex k. If we denote them by a(k) and b{k) one can easily show that they satisfy the relations a(— k) = — b(k), a*(k) = — b(k) (in taking the conjugate complex one must not change k to k*). We write the asymptotic form of y in the form X = A (e- ikr -^ — e ikr +*a) = _ 21 A sin (kr -+- 8 ). The phase shift 8 follows from the equation e * ih = J** (2*.) r(2rti+D ._ 4afci ln ax A bound state corresponds to an imaginary value of k = ik n and a negative value of the energy. For k n > o the coefficient of e~ ikr = *V in the first term of the asymptotic expression for x must vanish. That means that either J 2a * n (2ax) = 0, or r {2 J , t) = 0- From the second condition we get 2ofe B -Hl= — n (/i = 0, 1, 2, ...), or, E = — ^= fi2(* + l) 2 However, the order of the Bessel functions now becomes integral and since ,/„(*) = (— i)~y_ B ( x ), the two solutions are linearly depen- dent and the wave function (1) vanishes identically. The energy levels are thus fictitious. The first condition gives us the true discrete spectrum, / 2 «* n (2ax) = 0, E n = —~?. (2) The zeros of the expression e 2i5 »w lie thus on the imaginary axes, and apart from the values ik n , corresponding to the discrete 338 ANSWERS AND SOLUTIONS spectrum (2), contain also the redundant zeros. 8 - al da ( g2 \ 2 dQ c- n ^ gi ■i aa — \2EJ / . 2 » , n^Y ' A 2 * 3 (2(*£ + ft 3 * 3 ) ' V 2 ' 2jx£/ 4fi 2 a*£ 16|A 2 £/ 2 « 2 C) rfa = T7 7-; . , , 2 \4 ^' 64ti jj.2^0 1664 + 12#3g2 _[- 3g4 ° — 3 h* a* (a^ + 4^)3 * 9. a) We evaluate the atomic form factor for hydrogen, f If iqr • 1 where a =-^ is the Bohr radius. We find thus for the differential cross-section and for the total cross-section _ Tcq2 TkW + 18fe2fl 2 + 12 In this case the Born approximation can be applied, provided ka^> 1, so that we can simplify the last expression to b) For the helium atom we get from the variational method the following expression for the electron density distribution, 2 -x a 16 n{r) = ^e » , b= m a. The differential and total cross-sections for elastic scattering by a helium atom have in this approximation the same form as for SCATTERING 339 hydrogen. We must only replace a by b in equations (1) and (2), and introduce a factor Z 2 =4 . In particular, _2&t 10. The wave function of a system of two identical particles will be the product of an orbital and a spin function. Independent of whether the spin of the particles is integer or half -integer, an even total spin corresponds to a symmetric orbital wave function, and an odd total spin to an antisymmetric one. We can introduce the centre of mass system and separate the centre of mass variable; the orbital wave function will then be of the form V(r v r 2 ) = <p(/?)<],(p), with D r l + r 2 . -. „ K-- 2 » P — r i — r 2- If we interchange r t and r 2 the function <p(K), which describes the centre of mass motion, does clearly not change. The wave function of the relative motion of the two particles will thus be even, <Mp) = <K— p) if the total spin S is even, and odd, <Mp)=— <K— p), if S is odd. The unperturbed wave function can be written as <j> (p) == g**op _(- e -i*oP , (1) for even 5 and as ^ (P) = ^ i * op — e ~ **"f , (2) for odd S» The interaction between the particles produces a scattered wave -£-* <?**?, where is the angle between k and the direction in which the particles disappear in the centre Of mass system. The scattering amplitude can be expressed in terms of the scattering amplitude for a particle with a mass equal to the re- duced mass of the two particles in the field U(r). Indeed, <j>(p) satisfies the equation: 340 ANSWERS AND SOLUTIONS {_£(A p + A») — l/(p)}*(p) = 0. If the incident wave e ih °? corresponds to a scattered wave ^-J e ik ?, we get for the incident wave (1) ^oW = /<»)+/(« — ») ( even s P in )» and for the wave (2) / 7 1 (») = /(»)— /(* — ») (odd spin). The probability that one of the particles is scattered into a solid angle dQ (the other particle moves in the opposite direction) is re- lated to the density of the incident beam, <fao = |/(») + /(« — »)l*. *»i = l/(»)— /(* — »)!*. The amplitude /(()) can be expressed in terms of the phase shifts 8j as follows, oo /( d )=42< 2/+1)P » (cosft)I ' ltt, ~~ 11 - 1=0 Taking into account that P x (cos (tc — d) ) = P, (— cos 0) = (— 1 ) l P x (cos »), we get F o (n)=s 4 H (2i+l)P,(cosft)[e" 8 i_.i], even? Fl(, ' )== 7i S (2/+l)^(cos0)[^_lj. odd Z For slow particles, small values of I will give the major contribu- tion to the scattering. If the total spin is even, the cross-section is spherically symmetric (as is the case for the scattering of dif- ferent particles) and does not vanish as k ->■ 0. If the total spin is odd, the scattering is determined by the term with 1=1. Since h t ^k 2l+1 for small wave numbers k, the SCATTERING 341 cross-section vanishes as £ 2 (as £-► 0)and has an angular depend- ence — cos 2 0. 11. In the case of the Coulomb field the scattering amplitude, in Coulomb units, will be 2*2 sin2 1. "(-*)' Using the results of the preceding problem we find for the differ- ential cross-section for the case of even spin d°o = |/( n ) + /(rc — »)| a d2 = ~4k* I T"T »H A 5— > rf2 - sln^ cos<-g- sin 8 1- cos» y J This formula gives the cross-section for the scattering of a- particles whose spin is zero. In the case of two electrons we can have a state of total spin 1. The differential cross-section is then <*»i = l/(»)— /(*—») I 2 = . . 2 cos cos 4 -^- sin^ -~- cos 2 -^- If the scattered electrons are not polarised there are three possible values for the z-component of the total spin, S z = 0, S z = ± 1, and two possible values for the total spin, S = and S= 1. The prob- abilities for each of the possible values of the z-component are W_ x = \ , W = 1 , W +l = 1 . The values S 2 = ± 1 correspond cer- tainly to a total spin S = 1. Since the different values of the z- component for S= 1 are equally probable, the probability for S z = S = 1 is the same as for S z = ± 1, S = 1, that is, ^-. We find thus that the probability for total spin S = is w — ~ = — , and for total spin S = 1 equal to — . For an unpolarised electron beam we get thus 342 ANSWERS AND SOLUTIONS do = T rfa H--jrf3 1 = cos 4 -~- sin 2 -^- cos 2 -^- The last, interference, term inside the braces is characteristic for the scattering of identical particles. As fi->0 the equation for da must go over to the classical Rutherford formula which in the centre of mass system is of the form 4 * Sln^y cos* -^ e 2 The transition to this formula is somewhat unusual. If ^» 1, so that we can apply classical considerations, the interference term in the usual units is of the form cos ,0 - • sin 2 -jj- cos 2 -j and oscillates fast. The quantum mechanical differential cross- section is thus essentially different from the classical one even for large values of £ if » is strictly fixed. However, if we aver- age over a small range of angles A&~5 the interference term vanishes, and the quantum mechanical equation goes over into the classical one. 12 - ;« = |/;+.t/;+t^-^ ( ^- ' The average value of the operator («>;> in the state characterised by the spin function e~* a cos3 e ia sin 8 Lvov is equal to (a n a p ) = cos 2 B — sin 2 3. SCATTERING 343 We find thus for the scattering cross-section the expression ° = * { V\ + f\ — (/f — /pcos 23 } or, (3 , 1 cos 2p , v i = \7°» +4"^ _E-( 3 , — o,)j. In the case where the neutron beam is unpolarised cos 2£ = and we get for the cross-section 3.1 13. The spin states of the neutron and the proton before the interaction are described by the function 1\ /0 o/„\i /p This function can be expanded in terms of the spin functions of the singlet and triplet states, + 7jIM,-G).0J- The scattered wave is of the form or It follows thus that the probability for reorientation is equal to 1 (A -A) 2 2 4 + /? ' 14. We introduce the operator of the total spin of two protons, 344 ANSWERS AND SOLUTIONS One can easily show that Using this expression we get for / 2 f 2 = T {(A + 3/s) 2 + (5/5-2/,/ 4 -3/& (a n S) -f (/ 3 -A)2 52 } . For the case where the scattering takes place in the para-state the cross-section is equal to aP ara =*(/i-h3/ 3 )*. This cross-section does not, of course, depend on the polarisation of the incident neutrons since there is no preferential direction in space. The scattering cross-section for ortho -hydrogen is equal to o ortho = *{(/i "h3/ 3 ) 2 + ■4-C5/I — 2/.A — 3/f) cos 2p + 2(/,— A) 2 }. where 2p is the angle between the direction of the total spin of the two protons and the spin of the neutron. If the neutron beam is unpolarised the average value of cos 2p taken over a mixed ensemble is equal to zero and is of the form O ortho = ^ { (A+ 3/s)2 + 2 (/3-/J2 } , while the ratio of the cross-sections is equal to gOrtho Tpara 1+2 VA + 3/3/ 15. The radial functions which satisfy the boundary condition Xi(a) = 0, can be expressed as follows in terms of Bessel functions: Xl = YT {y_ r - v , (ka) 7i + , A (kr) — Ji + i h (ka) J-i-y t (kr)} . From the asymptotic behaviour of the Bessel functions we find for the phase shifts J 1 1/ (ka) We get thus for the total electrostatic scattering cross-section SCATTERING 345 17. At large distances from the target (the target consists of scalar particles) the wave function of the incident particles will be of the form <^(J)«SF.S'' +, < 2, + , >(J) P '< c0,,, >X 1 = We shall expand the function ( J) />,(cos») in terms of the eigen- functions of the operator A The result is (i)P,(cos6 )==1 ^_(l) rio(0)== = §q^i {VT+Tvt + vTvrj . (2) In this equation W z + and Tf are the Pauli functions (see problem 20, section 4), V5T+1 \- yT+TrJ V 2 ' l ' J * 2) • Substituting expression (2) into equation (1) we get x {,-(-!) _/(-3). 1=0 (3) The interaction changes only the outgoing wave ~ . Since for any interaction law >, /* and j x will be integrals of motion (see problem 39 of Section 4) this change will in general be different for 346 ANSWERS AND SOLUTIONS states with different quantum numbers j, l. We get the following expression for the scattered wave: W a 1 = or ijr 8 ~"7~ /* 2d V2T+7 x \Q Y n w+ d(v - o+'cv - l >i+ K «i<v— v)}- It is thus clear that the re -orientation of the particle spin can take place when ^ ¥= V We can express the scattering cross-section in term s of t£ and % • The differential cross-section for scattering with a simultan- eous change in polarisation, d 0l , is equal to da t = » z=o d£, while the cross-section, da 2 , without change in polarisation is equal to da 2=P ? = srw+l yioK'+ixv— 1 >+'(v — *>} dQ. If the relative velocity of the particles is not large we need take into account only scattering of the S- and P -waves, (K + — i|<ci. hr — H<o, In that case we have if />!)• SCATTERING 347 From the expression for rfo, it is clear that particles whose spin orientation is changed are only weakly scattered in the direction perpendicular to the z-axis. 19. Since /?>!, we can apply semi -classical considerations. R All particles will be incident with / < £ , so that A ■»)i = if /< R_ X' If we substitute these values of t\i into the expression for the total cross -sections, :*A 2 2(2/+1)(1-K|2), a 8 = ,rX 2 2(2/+l)|l-^|2, we get: % Z = We see that the total cross-section a = o r +o s is equal to twice the geometric cross-section of the nucleus. 20. In all three cases the distribution is isotropic. 21. The eigenfunctions of the operator I g for the nucleon-pion system can be written in the form of all possible products of the functions 9 and $. The functions ? correspond to the different charge states of the pion (<p + , cp , ? _) and the functions -> to those of the nucleon (.jy -> w ). We have given all possible combinations in the following table: 7=1 J z 2 '.-4 ("°)=-?o'l'» In this table (/?+) indicates the function of the system consisting of a positive pion and a proton, (n°) that of a system consisting of a 348 ANSWERS AND SOLUTIONS neutral pion and a neutron, and so on. These functions will, gen- erally speaking, not be eigenfunctions of the operator of the square of the total isotopic spin of the system P. The eigenfunctions of I 2 , which are at the same time eigen functions of I g , will be linear combinations of the functions from the table, multiplied by Clebsch-Gordan coefficients. The Clebsch- Gordan coefficients for m , = ± 1 l 2 are of the form (see problem 20, section 4). «' 1 m = T 1 m ==—2" ' = /+* V 2/4-1 f 2/ + 1 "J-i |/ J-M + Jf f 2y + l |/; + m + -1 r 2/+i In our case M = I g , j = 1 , mf = v If we use this table we get the eigenfunctions $/ of the oper- ators / 2 and l z : &£ = (P + ), ^=-/"t(/»-)+/"t(»°). *s-=-/"T(p°)+v r 4(* + ). One can easily express the eigenfunctions of the nucleon-pion system in terms of these eigenfunctions: (p») = /"I •&-/£ •& (»°) -/f *%,+ /T *** . SCATTERING 22. We can expand the incident wave as follows: * = * a '(J) 8 (* — *«)*(» — \> = =2^(2/+l)Pi(cos6)(J)8(«_ic08(»-^^^» 349 z=o J=0 J (1) where cfy are the Clebsch-Gordan coefficients which are given in the following table 1 X *=2 ■'. = — J c l VWt /l-'f z -fi-'t /±+' f (2) We have used here (see preceding problem) 8 (ic — *i) 8 {n — i e ) = 2 C?«$ f . r « * We introduce the Pauli functions for mj = V 2 (see problem 17, Section 9): Yt -|_ i Y io and expand ^ ( q ) ta term s of ^^ : Substituting expression (3) into equation (1) we get (3) * = ^ S S '» cVJ+Tk? + VTVf 1=0 I )X 350 ANSWERS AND SOLUTIONS xs,-„(* r -f)cjv^ = OO z=o J XCjV^f (^ — (—l) 1 *-*"")=* 2 z = ^22 <*>**. (i / '+ TK '' + "< Kr > ^ - OO X Z.---0 / Xi—W-r-- ( 4 ) 23. We shall use the expansion of the incident wave in terms of the eigenfunctions of the operators which are conserved (see equation (4) of the preceding problem). Each term of this sum, which corresponds to well-defined values of I, J, and /, will be scattered independently of the others. The number of particles corresponding to given values of I, J, and / will thus not be changed during the elastic scattering pro- cess, so that the influence of the scattering centre will be to intro- duce some phase factor e 2t * l± . The quantity Sj+ =o i depends on I, J, and / but it is independent of I t because of the hypothesis of isotopic in variance. We must note that the scattering centre does not influence the incoming wave but only the outgoing one. The wave function of the system can thus be written, if we take scattering into account, in the form CO / Z = OO -^22^ on+TYt+VTrn'-l <D i 1 = SCATTERING 35 * At large distances from the scattering centre we can write it in the form ,e*kr 6 = 4. +/— . * 'lie ' J r The quantity f is the scattering amplitude. If we take from equa- tion (1) the expression for ^ c (see equation (4) of the preceding problem) we get : I l-o X *?, {VT+TYt {e 2i ^ - l)+ V7l7 (•**- - l)|. (2) The functions #f # can be expanded in terms of the eigenfunctions of the operator /, <&/ g = c£*8(*— *<)&(*— Tj+C/^-JCic — **)&(» + *,), (3) where the Clebsch-Gordan coefficients c]]* are given in table (2) of the preceding problem and where * k = w, + 2 v Substituting equation (3) into equation (2) we get: I I«0 x {yT+rvr (/* - O+ynr (•■"*- - 01. where and t\ denotes the final state of the nucleon. If we now put /=/**8(ir — n^lin — x g ) + Z +/« &(«—*»)&(» + «■). 352 we get: ANSWERS AND SOLUTIONS +V7it(."<--i)). where G Ix », is given in the following table for all reactions necessary for our problem: (4) Reaction /?+->/>+ P '-+P~ p -*/l« 1 1 3 3 2 3 Y2 3 If we substitute these values of O u ', into equation (4) we get finally /(P+ , „ +) _£ |{v7+7r,+(< 2 * - i)+ /(/>-. p-) — ^ S (l^+Tn + (."* +2."* - s)+ _ '"' + V/K,-(^ + 2 e 2 *-3)}, | OO HP-, O - ^ 2{vTTTrt (/"*-#"*)+ r»o (5) 24. The table. of the coefficients G*\ for all possible reac- tions of the pions with nucleons is of the form SCATTERING 353 No. Reaction 4 A > 0*> 1 | />+-/>+ 3/2 1 2 | /*>-/*> 1/2 2'3 1/3 3 />° -*•/! + 1/2 V2/3 - V2/3 4 p~ -*-n0 -1/2 | V2/3 — V"273 5 p~-+p- -1/2 1/3 2/3 6 /!+-»> rt + 1/2 1/3 2/3 7 «°->/?~ -1/2 ^2/3 - V2/3 8 /!+-». /*> 1/2 ^2/3 - V2/3 9 1 1T| /!» -*/!<> -1/2 2/3 1/3 n~->/i~ — 3/2 1 (1> Since the phases are independent of /, because of our hyoo- ttesis of isotopic invariance it follows immediately from equaTon (4) of the preceding problem and table (1) that <=4"«"*i 1) 2) 3) 4) f(jfi, «+)=/(«+, Jfi)=f { p-, ft )==/(rt0> p.), The expressions for the first three amplitudes were riven in th* w „ ceding problem. From table (1) one sees ea^Uy Sat ^ - Using table (1) we get /(/>+. p+) = f(n-, «-)=/%, 354 ANSWERS AND SOLUTIONS =/(/>-. *°)=/(«°. /r)—J^I/y— /«/•]. 4)/(p°. />°)=/(* , «?)--5[2/ , M-/*l. 25. The differential scattering cross-section is equal to ^==|/|2, where dQ = sin8d8d<p. The total scattering cross-section is equal to o = J* f |/| 2 sin6d0dcp. If we substitute here the scattering amplitudes for the reactions considered, which are given in problem 23, sections, and use the orthonormality of the Pauli functions / j* (Yt) + (Xr) d® = n (r f)+ (F ,7) dQ = 8 "' ' j f(Yt) + (Yl)dQ = f f{Y{) + (Yt)dQ=0, we get oo «(p + .p + )=FSi( / + 1)sin2 ^ +/sin28 ; / -]- 1=0 X [sin* 8V. 4- 2 sin* 8* - -| sin 2 (8 3 /. - 8V;)J + 4-/ [sin* r/i -f 2 sin* 8;/ L — |- sin* (8£ — 8^)]} , a(p-, «°)=|^£((/H-l)sinH^-W-h^in*(8;/ 1 -8;' 1 )). 1 = + 26. We shall give the detailed solution for the reaction (p P + ). From problem 23, section ft. we get for the scattering ampli- tude for S- and p-waves where a = e ° — 1; ai=« 1+ — l, Pi — « l - SCATTERING 355 Using the explicit expression for the Pauli functions of problem 22, Section 9, we get for the differential scattering cross-section for S-and P-waves 7rl Fool I *\ol {2(«*!+ o^x 1 )+(«o?i + «oPi)} + ^V3 -fyino! 2 i4|a 1 | 2 +|?i| 2 +2(a 1 ^-i-aM} + + fl^| 2 i!« 1 | 2 + IPi| 2 -(^ + a: i 3;)) Since the spherical harmonics Y lm are equal to we can write the differential cross-section in the form, k 2 ^ = A + B cos 6 -f- C cos 2 6, where the coefficients 4, 5, and C. axe given by (1) fl = l {2(00^-1- a^i) -h- («o?T -h «SPx)} : C-ltaf + fotf + afo)}. To express the coefficients A, B, and C in terms of the phase shifts we use the identities I e *ix— 1 p = 4 sin 2 *; = 4 (sin 2 jc-j- sin 2 ,y — sin 2 (x — y)]. We get then A(p+, />+)=sin 2 8V.+sin 2 (8% — 8%); B (p + , p+) = 3 sin 2 8'/.-|- 2 sin 2 &;/•_ -+- sin 2 8^ — — 2 sin 2 (8V. — 8V;) — sin 2 (8;/. — 8*); C (/>+, />+) =» 3 (2 sin 2 8V; + sin 2 8% _ sin 2 (8% — 8Vi)) , (2) 3S6 ANSWERS AND SOLUTIONS We can easily extend our method to find the coefficients A, B, C for the reactions (p-, p -) and (p-, n°). We shall only give the final results: HP-, />-)=4 sin 2 8; / '+|-sin 2 8^-l S i„ 2 (8;/'^8^)4. + Isin 2 (8ft_Sft)-|- s in 2 (8ft-8ft)4- + |sin 2 (8£ _8ft)+ 2 S in 2 (Sft _ e) + + |-sin 2 (Sft -8ft)- * sin 2 (8ft _8ft ); £(p-, />-) = sin 2 8o /j -f 2 sin 2 8o /f + J-sin 2 8ft + + 4sin 2 8ft-h-isin 2 8; / !. + |-sm 2 8Vi-|sin 2 (8; / «-8^)_ -i-sin 2 (8^_8V;)-4-sin 2 (8^-8Vl)- -l-sln 1 ^-^)-^ 1 ^-^.); C(/>-, p-) = 2sin 2 8ft + 4sin 2 8ft + sin 2 8ft-f + 2sin 2 8ft_2 sin 2 (8ft_8ft)_ -i-sin 2 (8ft -8;/l)_| sin 2 (8ft -8ft) - -lsin 2 (8ft -8ft)_ 2 sin 2(8ft_ 8 ft); (3> A{p- t »o) = -| {sin 2 (8^-^) + sin 2 (8ft _ 8ft) + -h sin 2 (8ft - 8ft) - sin 2 (8ft - 8ft) - -sin 2 (8ft_8ft) + S in 2 (8ft-8ft) + + sin 2 (8ft-8ft)}; SCATTERING 357 B( P -, « ) = |{-2sin 2 (8V.-8:y_ h 2sin 2 (8^_8;y_ - sin 2 (#_ 8&) + sin 2 (ft- 8*.) + + 2 sin 2 (8* _ jv. ) — 2 sin 2 (8* - 8* ) + -f- sin 2 (8* _ ft_) _ sin 2 ( 8 V. _ 8 y L )| . -^'(^-ft)H-«In«(8K.-.8;y-d ll «(8a-8}!L)}. (4) „* A 1 l S obtamed are of «^at importance for the evaluation of the angular distribution for the scattering of pions by protons. We can verify experimentally equation (1) and determine the coeffi- cients UttdC for the reactions (p+ , p+) , (p - t p - )t ^ % \v* *v * We I/ can I then ^ermine the six unknown phase shifts 80', 8 \ 8/;. 8/i, 8* , 8['L from the equations <2), (3) and (4). However, the evaluation of the phase shifts possesses an ambi- guity: first of all because of the fact that in equations (2), (3) and <4) the square of the sines of the phases and their differences enter so that we cannot determine their sign; secondly there are se^S different sets of phases satisfying the experimental data. Out of tiiem the solution by Fermi in which the largest contribution to the scattering comes from the phase ft gives the best agreementwith experiments. This phase corresponds to a scattering over 90" for a pion energy E^ 195 MeV in the laboratory system; the other phases ft_, ft, ft_ ar e small. There are a number of other criteria which can be used to get rid of this ambiguity. The signs of the phases can be determined from considerations based on the causality principle, and also from experiments which take the Coulomb interaction especially into account. In choosing the correct solution one might be helped by an experiment about the polarisation of the recoil nucleons but this experiment has not as yet been performed. The expected value for the polarisation in the reactions („ < , p .), <„-, p - )t and ( P -, n <>) are determined in the following problem. 358 ANSWERS AND SOLUTIONS 27. We consider in detail the reaction (p+, p+) a== (o) if s * = 1/2; P == (l) if S ^- 1/2 ' If originally we had *, = »/.. ** the proton, we can write the scat- tering amplitude in the form (1) While, if initially *, = — »/, we have /-*=/?««+/???• (2) In these equations /„ and / P{J are the scattering amplitudes with- out spin re-orientation and f a? and /„ those with spin re-onenta- tion. The amplitudes /. and /„ can be deterained ^ om ft ^ a " tion (5) of problem 23, Section 9, as the coefficients of the columns ( l ) and ( J ) • If we take only S- and P-waves into account we get: / M = lI|a K 00 +^=[2a 1 + ? 1 ] K 10 J. (3) where «„«,?, were given in the preceding problem. If we use the expressions for the Pauli functions with «,-»»/* we fi* deasll y If we assume that the pions are scattered in the **-plane, the polar angle <p = and we have thus /«9 = h*' SCATTERING 359 We get then for the scattering amplitude for a proton with * = - v (see equation (2) ) z ' 2 f- v* = — /p« a H- /««? • (5) Since the protons were originally unpolarised it follows that after the scattering they are still unpolarised along the z-axis. One can show that after the scattering there is no polarisation in the xz -plane. Indeed the functions of the proton Te and 8 6 , which correspond to +£ and 4 components of the spin along a z' axis which in the xz -plane makes an angle ft with the z-axis, will be of the form (see problem 19, Section 4) We get thus /i/ t = =5 /«« a 4-/«pp = = (/„ cos j - /.„ sin i) T9 + (/_ sin | + /^ cos 1) 8 8 ; = -(/ aa sini- + / op cos}) T9 -f(/ M cosi--/ ap sin|)8„ which means that there is no polarisation in any direction in the xz -plane. The protons will, however, be polarised along the y-axis which is perpendicular to the plane of scattering. To find the mag- nitude of the polarisation, we express / and / in terms of the spin eigenfunctions k which correspond to a spin direction parallel or anti -parallel to the y -axis. We get then 360 ANSWERS AND SOLUTIONS It follows that 7 +~ I /«—//., I 2 ; 'W r -~l/„+//. p | a . (6) r where w + and W_ are the probabilities that the spin after the scat- tering will be parallel or anti -parallel to the y-axis. We note that equation (6) is valid independent of the initial values s g of the proton. If we substitute into equation (6) expressions (3) and (4) for / tta and f a? , we get: W±~ |« 4-(2a 1 -|-? 1 )coseH=/(? 1 — ai )sin8|2 or W, | (,»#_ ,) + ( 2 , 8tt & _3 +/<-)cos 6 ± We find similarly for the reactions (p-, p-)and (p-, n°): V±<p-.p-)~i(.^ , -3+2. t< #) + + ( 2 ,'«&_9+V^ + « 14 #- + 2* 2i8 *-)cos 8±: W ± (p~, n°)~\{e U ^-e 2i ^) + +- (2. 1 * _ 2/ **• + /<'- _ g * <#-) cos e h- . ./' 2t8?' 2i8V» 2i8*/» . 2i8 1 /* \ . J 2 28. * It is well-known that the solution of the problem of elastic scattering by a potential U(r) in the Born approximation leads to the following expression for the wave function, ,{,~<J,(0)_|_»J,(1), (1) where y® (r) = e ihr ' t \ ik\ r-r>\ 1 (2) r(r) = __^_J f ;(r')f0)(r04^1 i ^' SCATTERING 361 so that asymptotically we have where q = k— k f is the change in the wave vector of the particle dur- ing the scattering, and where 6 is the angle of scattering. The differential cross-section for scattering into a solid angle- dQ over an angle 6 with the direction k of the incoming particle is equal to m a do = |/(e)p d2 « ^jl I Vir>)eWdr<\ dQ. (3) We shall evaluate the integral in equation (3). Taking the polar axis of our system of spherical polars along the vector q and substitute for £/(/•'), we get: J e - %r '+ iqr '^^\e-"' r> dr' le^dx^^, and thus da — ( 2Am V <& ? = 2*sinl. (4) Integrating expression (4) over the angles we get the total scat- tering cross-section 4w (2AmV 4* ( 2Am V *?-}-- 4)K *fia ) ' v>) Let us consider some limiting cases of equation (4). (1) k <C x, so that also ^<x: **:(§£;*. (4-) that is, the scattering cross-section, in complete accordance with the general theory of scattering of slow particles, depends neither on the scattering angle nor on the particle energy. 362 ANSWERS AND SOLUTIONS (2) k ^> x (fast particles). There are now two characteristic ranges of scattering angles, a) 6 <C j , or q <: x; the cross-section for these small angles of scattering tends for G-+0 to the constant limit (4 f ); b) 8 :> j , or , q :> x (large deflection). In this case we have, introducing the particle energy E==^ , SIn4 2 This is essentially the Rutherford scattering formula, as it should 1 be, since in the essential region of distances, r ~~z the fie ld is practically a Coulomb field. The condition of applicability of the Born formulae (3) and (4) is that the correction term ^ (1) , which describes the influence of the scattering field, is small compared to the main term <J> (0) . This condition is always satisfied, if I V 1 ' | r , <C |<|> (0) | = 1. This condition of applicability is sufficient; it can, however, be shown to be too strong, especially for small scattering angles. We have oo B _ 2HL± f e -rr'+ikr' sln kr ' dr \ (2 n ) «= p-J * kr > * ' For slow particles (k^.*) the important region in the integral is r ' ~ - , where e ikr ' --^r-'- ~ 1 , so that the integral is of the order x kr of magnitude of — and the required condition of applicability of equation (4) is of the form ^<1- (6«) SCATTERING 363 For fast particles (ft ^> x) the presence of the fast oscillat- ing function {e ikr " sin kr r ) means that the important region is r'~l-, so that the integral is of the order of magnitude of -j~ . and the con- dition of applicability of the Born equations will be (introducing the particle velocity v = ~) : The meaning of inequalities (6*) and (6") becomes very obvious if we consider them in conjunction with the limiting cases of expres- sion (5) for the total cross-section. Dropping numerical coefficients we see then that both inequalities are equivalent to the condition The condition of applicability of the Born approximation is thus that the scattering cross-section is small compared to the square of the effective range of the scattering field, or, in other words, that the scattering amplitude YH is small compared to the range of the field - . Using the results of this problem we can consider as a parti- cular case the scattering of electrons by neutral atoms (that is, by a screened Coulomb field). We must then replace A by — Ze 1 and - by the Thomas-Fermi radius of the atom which is of the order of magnitude of m Z l/s me 2 ' 29. * The scattering potential, or rather potential energy, is equal to t/(r) = £§(/•), (1) where we have taken the force centre as the origin. The constant B is clearly equal to the volume integral of the potential, B= ft/ (r)dV = const. (2) 364 ANSWERS AND SOLUTIONS The differential cross-section for elastic scattering into a unit solid angle (in the centre of mass system) is equal to d<3 do ~ 4it2/j4 & | \u(r)e«*r d v\, (3) where n is the reduced mass of the colliding particles and hq~s eh/j — p' is the change in momentum of their relative motion. From equations (1) and (3) and the properties of 5(r), we get: do ~ 4%W The scattering by a delta -function potential is thus isotropic and does not depend on the velocity. It is well-known that the same properties characterise the scattering of sufficiently slow particles by a potential well of finite dimensions. We shall elucidate the con- nection between these facts shortly. The total scattering cross-section is equal to 0==4 *£«i^-. (5) do llfi 4 If we bear in mind the meaning of the constant B [B = J U dV) , we see that equations (4) and (5) are obtained from the general Born equation (3) always when e iqreii « 1 , or | qr eft | <C 1 . In order that this will be the case for all scattering angles, it is necessary (since g max ~k) that ftr e{{ <l or r eff <:X> where r e ff is of the order of the dimensions of the region where U(r) is appreciably different from zero (" range of the interaction"). The delta-function potential considered by us is thus an ideal- ised interaction potential with a very small range, or rather, with a range which is much less than the de Broglie wave length of the relative motion of the colliding particles. The delta -function pot- ential can thus, for instance, serve to describe formally the inter- action of sufficiently slow neutrons with protons or very heavy nuclei. In those cases the formal application of the Born approximation leads to the correct result, even though perturbation theory is es- sentially inapplicable to a delta -function potential of the form con- sidered by us. Indeed, the condition of applicability of perturba- tion theory for slow particles (ka^.1) is of the form ^l<C^r a . |*fl2 SCATTERING 365 If we write this condition as | U | a» <^ _$! a and bear in mind that for our delta -function potential \u\a* has a fixed, finite value, we see that for sufficiently small values of a (and always if a = 0) the perturbation theory, and thus also the Born approximation, is in- applicable. From equations (4) and (5) it is also clear that the formal ap- plication of the Born approximation leads to the correct result in the case where the true cross-section does not depend on the velocity. In this case we are, roughly speaking, not interested in evaluating the scattering cross-section for the given potential, but rather in "adjusting" the constant of the potential to the known value of the cross-section. 30. * The Schr5dinger equation of our problem is of the form Introducing, as usual, k* = Qf we get this equation in the form ^ + ^ = •1^. (1) We shall consider the right hand side as an inhomogeneous term. We must thus find a solution of an inhomogeneous equation, satisfying a given boundary condition, tj> rv e iKr -I / ****" r-*oo f Using the expression for the Green function of equation (1), /-./ /x 1 e ik fr-r ' I G{r — r') = e 4k \ r — r'\' we easily find the required solution For large values of r we have where Hence, e ik\ r-r' \ Akr , — . K - i \r-r'\ ,~ r -C r ikr' if -> OO), 366 ANSWERS AND SOLUTIONS 4 « e i*,r — ^_ •£ J dVe-^U (/) <j> (#0. so that '<*> " — 2&J dVre_ifcrt/ W * ( r >- This equation is convenient for various approximate calcula- tions. For instance, by substituting <|>(r)«e**° r we get the scattering amplitude in the first Born approximation. 31. * We take the origin in the centre of rotation and write the potential energy of the interaction of the incident particle with the rotator in the form ^ = 17-4^1' (1 > \r — na\ where e is the charge of both particles, a the distance of the rot- ating particle from the origin, » a unit vector along the "axis" of the rotator, and r the radius vector of the scattered particle. The probability per unit time for the transition under consider- ation is in first order perturbation theory equal to dw 0l = ^\U 0l \2p E dQ, (2) where U ol is the matrix element of the perturbation (1) for the transition considered, taken between normalised wave functions; p E aQ is the number of final states per unit energy, where dQ is an element of solid angle around the scattering direction. The initial and final wave functions of the system are of the form 1 1 %H T ^ in =y=e^r Y(X> (n), <I> fin =^'K ta («), ( 3 ) where the Y lm are normalised spherical harmonics describing stationary states of the rotator, k Q and k T axe the wave vectors of the scattered particle before and after the scattering, and V is a normalising volume. From equations (1) and (3) we get: U lo =U*oi= J j ffin ^in drdo = = e ;^YUn)Y 00 (n)e^-^ T ^Jl rr (4) SCATTERING 367 where do is an element of solid angle around ». From the calculations it will become clear that for any given I only one state of the rotator (with m = 0) can be excited. The density of final states p^rfQ is thus solely determined by the num- ber of states of the scattered particle and hence equal to itfg* P^ Q== Vnhf ' (5) where p x = hk x is the momentum of the scattered particle which is related to the initial momentum through the law of conservation of energy e _p\_p\ , fi 2 '('+D C ~2 ( *~"2 f x" i 27 ' l 6 > where ft is the mass of the scattered particle and / the moment of inertia of the rotator. If we write k Q —k l = q and use the fact that K 00 = — L and change y 4n the variables to r — na = r' we get instead of equation (4) The Fourier component of the Coulomb potential [e^' — , which occurs here is equal to -^ (see, for instance problem 33, Section 3). If we choose q as the axis for quantising the rotator and expand <?*'««" in a series of Legendre polynomials we get: X J J CXO, <p)/V(cos0)sin8tf0rf<p. Writing Y* lm =y r2 -i±^-.^ r ^ l PT(cos^e-^ > we see that only those matrix elements are different from zero for which the transition is it to a state with t» = 0. Using the relation f P X P V sin tf& = -- 2 ... 8 ,,, and '. summing over /', we get for those matrix elements the expression Ult = U l0 = W ^ |/"(2/+D^ J l+L (ga). (8) 368 ANSWERS AND SOLUTIONS The differential cross-section <*o,(6) is equal to the probability (2) divided by the current density of the incident particles. The latter is according to equation (3) equal to y. Moreover, the quantity dp, 1 d(ph . ... . .. ftjE ~2 Pl ~JE~ o 001 ^ 11 ^ m expression (5) is according to equation (6) equal to f*p, = H. 2 v,» where v and i>, are the velocities of the scattered particle before and after the scattering process. Combining this result with equations (2), (5), and (8) we get finally rf3j(6 )=^ 01 =2.(2/+l)^%^ Q . (9, The cross-section does not contain the normalising volume V, which is as should be since it has no physical meaning. Equation (9) determines thus the required differential cross- section for the scattering of a particle into the element of solid angle dQ with simultaneous excitation of the l-th level of the rotator. The quantity q occurring in equation (9) depends both on the scattering angle & and on /, since p ==Pi-\-hq, where hq is the momentum transferred from the particle to the rotator during the collision. From the definition of q and equation (6) we get q* = kl -+- k) — 2k k t cos b = = 2^ — ^-/(7 -hU — 2fe |/'4— j'('-hl)cos8. (10) To find the angular distribution of the scattered particles irrespective of their energy, one must sum expression (9) over all values for I for a given value of 0. To find the total scattering cross - section for a given energy transfer, we must integrate expression (9) over the angles for a given value of /. Because the cross - section (9) depends in a rather complicated way on / and 9, both operations can in general be performed only numerically. However, for the most essential cases one can approximately give also an analytical evaluation, which we shall consider in the following. Let us now discuss equation (9); we note first of all that it is valid for the case where 1 = 0, that is, for elastic scattering. In this case q = 2£ sin -i, so tnat the elastic scattering is equal to ?*(* 4 (qa) SCATTERING 369 Wa< S,n2 ( 2 ^ Sln 4) * (2Msin|) 6 dQ, (11) where we have used the relation J, /l (qa)= y — sin^a. The total elastic scattering cross-section diverges clearly for small angles 0. In the limiting case where 2k a <C 1 expression (11) leads to that is, the Rutherford formula. This we should have expected since 2fc a <Ct 1 means that the dimensions of the region in which the scattering Coulomb centre (the rotator) moves, 2a is small compared to the de Broglie wave length of the scattered particle X = -? and this in turn is practically equivalent to having a fixed scattering centre, which is necessary for the validity of the Rutherford formula (12). For our next consideration it is convenient to transform expres- sion (9) slightly, going over from dQ to dq. From equation (10) we have for a given value of / qdq = k k l siT\QdV=*^dQ, (13) so that equation (9) leads to *.=(^)'< 2 '+'>%r^- <"> We shall limit our considerations to the excitation of the rota- 2 tor by a fast particle, whose initial energy £=- ^ is large com- pared to a "rotational quantum" ~ so that it follows from equation (6) that 0</</ max with l me ^> 1. (One verifies easily that for values of }x and / corresponding to the collisions of electrons - and a fortiori mesons or protons - with molecules the condition E ^> ^ ensures the validity of the necessary condition for the applicability of the perturbation theory used by us to scattering in a Coulomb field, ^;<C 1.) Furthermore, we make the approximation (confirmed by the final result) that the deciding part is played by collisions corresponding to small scattering angles (0<C! 1) and to a transfer of energy which is small compared to the energy of the incident 370 ANSWERS AND SOLUTIONS particle (A£<C E). In this case we have clearly k Q — &j<C k , so that A£s= §7 =,— (^ _^)^_0(^ _/5 i ) f (15) ^ = (k — ^ -f 2k k t ( 1 — cos 6) » (* - /^ -f- ftg6 2 . (16) Combining equations (15) and (16) we get .« /p^f+cw- /(^r+(w. <i7) AC For 9 = we have q~j— . and for A£ = we have q?ak ( p = = Tf. so that, roughly speaking, — and /? o are the longitudi- nally and transversely transferred momenta. It is clear from equation (17) that for not too small angles of scattering, for which k Q^> ~ , q depends only on 8, and not on I In that case we can sum expression (14) in its general form over / and thus obtain the angular distribution of the (both elastically and inelastically) scattered particles. We use the identity oo 2(2/ -f l)^(x)«7. (18) 1=0 2 (This identity can be obtained by integrating the absolute square of the expansion OO ,tooos8 ^ ^ (2/+ 1) i l y £J L (x) Pi (cos 6) over sin8rf8 and using the orthonormality relation * From equations (14) and (18) it follows that *w-f>,(,)-(g)r^-8.(£)r3. w 1=0 SCATTERING 371 Since for the angles considered at this moment qzxkffl, equation (19) can be written in the form da (6) =8* (19») which agrees with the Rutherford formula (12) (we remind ourselves that fl<Cl). This result is connected with the fact that for the angles considered P( p^>^, that is, the scattering is accompanied by the transfer of predominantly "transverse" momentum and is practically elastic. The range of scattering angles for which the angular distribution (19) or (19 1 ) is approximately valid is (where we have used the relation p v = 2E). Since we have assumed ( A£, )eff <C£, there can be a substantial range of angles satisfying con- dition (20). The character of the angular distribution (19») - a fast increase for decreasing 6 - confirms our earlier assumption that «eff<Cl. We must make the following observation. We have obtained equation (19) by summing over all values of I. However, inequality (20) and equation (17) impose on / the limit £? <C6 O 0, or /<C/, = / /ft — k oy j> which at first sight seem s to make equation (19) incor- rect. In actual fact, however it turns out that the condition /<C/ X does practically not limit the applicability of that equation in general. Indeed, one can easily check that for a given value of * the domi- nant role in equation (18) is played by the terms with /<*. It fol- lows that in order to satisfy the condition that equation (19) is appli- cable, leti^li, it is sufficient that qa<^,L or fl<c: — =3-^ (a — - fxa a |jt V rror ~ <x a is the mass of the rotating particle). In practically all cases of in- terest fJ rot ^n, so that the required inequality is automatically satis- fied for «<l. We shall now evaluate the total scattering cross-section with simultaneous excitation of the /-th level of the oscillator, o r . We integrate thereto expression (14) over g between limits correspond- ing to 6»»o and b = «. Since the main contribution to this cross- section corresponds to small scattering angles, we can use equation 372 ANSWERS AND SOLUTIONS (17) to find the limits of g and put = and ~ 1 . We have thus "max °«n 0, «mfn °«n "max "^raax with AE fl'mln — fa, Q • ^max ~ *0 ' (22) (to estimate the value of q ahx one may neglect the first term under the square root sign since A£<CE). While for large values of x the integrand in expression (21) de- creases steeply with increasing values of * for any I its behaviour for small values of x depends strongly on the value of I . Since J x (x)~x * + 2", as x -> the integrand will be proportional to x 2i-f> as * -* and thus tend to zero for /> 2 and to infinity for / = 0, 1. We shall consider these two cases separately. 1) 1 = 2, 3, 4,... In this case we have J V^Wx*- J J i + ± ix) T*~-T n -(iTW • °«mln (23) Let us first discuss the validity of replacing the limits of integra- tion by and oo. Since the first (largest) maximum, and also the first root, of the function J i (x) (where l^>\) is near x = l, while the function for jc<0 behaves like x 2 , and for x >l is heavily damped and oscillating, it is clear that the change in limits is al- lowed provided aq mia <^l> "Vmsx^ 1 - (24) Using equations (22) and (15) (to eliminate / ) and again the relation n = Z we can write these conditions in the form rrot fl 2 ' Since, as was mentioned earlier, usually t* rot ^t*-. the first condition is automatically satisfied for A£<£ but the second condition is SCATTERING 373 more restrictive. For the case j*~p rot at least we have thus proved the validity of the change in limits in expression (23). From equations (21) and (23) we have finally °'~!(0( 2/ -H>^fr™ 8 0-2. 3,...). (25) This total cross-section decreases steeply for increasing I (for /:>2as -, that is, as -L-), which also confirms our earlier assumption that (AE) eff <C E. 2) /=0, 1. The case / = o corresponds to elastic scattering and was con- sidered earlier. For /= 1 the integral (21) diverges logarithmically as x -+0, so that the main contribution to it comes from values of *<T1. If we therefore replace J vM by (*)* X ^ = ^L and integrate from fl ?mi n = "g^ = JxT^ 1 t0 some value aq~ I , we get *-*(rW«£)«" «=<>. (26, where c is a dimensionless quantity of order unity (its exact value is not very important since the argument of the logarithm is much larger than unity). The total inelastic scattering cross-section for all angles and all possible excitations of the rotator is finally equal to inel = °l + 2] °i = 1=2 -4(5^["(«£)+!<a+i>^]. where we have used equations (25) and (26). Using the relation (k + n-l)l (/r-2)(/ J -l)i» we get after some simple manipulations (/-2)i 1 1-2 S(B+I) £p£_J. 374 ANSWERS AND SOLUTIONS so that we finally get •n-f(£M*&)«* (27 > where c x =~cfl* (e = 2,718...). Since i-<Cl, oj»-i is small compared to na 2 which is the geometric cross-section of the region in which the rotator moves. The cross-section for excitation of the 1=1 level is much larger than the total cross-section for the ex- citation of all the higher levels. 32. * The two nuclear reactions are each other's inverse, or n+p^td + i. m (A) (B) ■* ' The principle of detailed balancing for the reactions (1) can be written as SbPb 3 b-*a SaPa (2) In equation (2) a A ^ 5 , a B + A are the cross-sections for the corres- ponding transitions, integrated over all directions of the velocities and summed over all spin directions of the final state, and averaged over those directions in the initial state: p A and p B are the momenta of the relative motion of the particles in the states A and B and g A gB the spin (and polarisation) statistical weights of these states. In our case we have ~o a + b=* <fea P i (capture cross-section), OB-*.4 = fr p h.diss (photo-dissociation cross-section). Since the neutron and proton spins are 5 and the deuteron spin 1, while the photon has two states of polarisation, we have ^ = (2-1+ O'^ 4 * ^ = (2-1 + 1). 2 = 6. We shall take our system of reference to be fixed in the deuteron. We have then p B =* P, = ~ where m is the frequency of the photon. One can easily verify that our system of reference is practically the same as the centre mass system of the states A and B . Indeed, in the exact centre of mass system we have by definition /> d -fp T = 0, or p d = P r = = *!! so that the deuteron velocity in this system of reference (and c * thus, conversely, the velocity of this system of reference relative to the deuteron) is equal to ^=^=^" c - Oa the other hand, ifthe photon SCATTERING 375 energy jfo>, is small compared to Mc*, and not too close to the deuteron binding energy 8 (we shall assume that these conditions are satisfied), the velocities of the separating nucleons will be of the order of v n ~v p ~ /l-^/"*. so that 54 ~ j/"g<: 1. The relative velocity of the two system s of reference is thus, indeed, small compared with the velocity of the products of the reaction and we may assume that both systems of reference are approximately the same, We have thus p 2 A ^2 i iEj. with n = ~ , where fi is the reduced mass of the n- P system, M the nucleon mass, and E A the energy of their relative motion (or, the energy in the centre of mass sys- tem). From the energy conservation law we have E A = E B , and clearly E B = hu> -+- (— a j = hta — e. We have thus p\ = M (tuo — e). Collecting all our results we get finally, °capt 3 fio> fi<o ff ph.dis72"^*^~' (3) We emphasise that this result is independent of any definite assumption made about the mechanism of the reaction, but is based solely on the reversibility of quantum mechanics (that is, on the symmetry of its equation with respect to time reversal). 33. * Let us note first of all that since the rest energy of a charged meson is larger than the rest mass of a neutral meson by about 5 MeV, while the maximum possible binding energy of the (ir + 4-7r~)-system is only about 2 keV. E = ~ 2P = - m • 2P" ~ ~ 2" ' 270 ' (13 " 5 eV) »- 1800eV. we do not consider possible here bound states produced by non- Coulomb interactions of the pions, the process under consideration is energetically possible, irrespective of the value of Z. The required selection rule arises because the two Tt -mesons , occurring in the final state, are identical. Indeed, since the pions have integer spin ( = 0) the (rc°-f u<>) system must be described by a wave function, symmetric in the two mesons. Since there is no spin -wave function (or rather, since the spin -wave function = 1), we must thus have a wave function which is symmetric in the coord- inates of the two it°-mesons. One can easily verify that the parity of the wave function of two particles 1 and 2 is the same as the symmetry of the coordinate wave function (the origin lies in the middle of the line 1 - 2); the wave function of the u° + it° system can 376 ANSWERS AND SOLUTIONS thus only be even(/ = 4- *)> and hence using the relation /== ( — l) 1 , for the parity of the orbital angular momentum wave function, we find that the orbital angular momentum of the relative motion of two it Hiiesons can also be only even. If we now take into consideration the conservation of the total (in our case this is simply the orbital) angular momentum, we con- clude that the reaction is possible for an initial state with even I but impossible for an initial state with odd I We note further that we cannot conclude from our considerations that the pion possesses negative intrinsic parity since both in the initial and in the final state there are two mesons, so that the total intrinsic parity has the trivial value + 1. 34. * We shall apply to this process the conservation laws for angular momentum and parity, and the Pauli principle. The total angular momentum / of the deuteron is equal to unity. If we take into account the assumed values of the spin and angular momentum of the % -meson the total angular momentum of the original nf-j-d- system is also equal to unity, and so is the total angular momentum of the final n -+- n -system . If we denote the orbital angular m om entum and spin of the two neutron system by L and S we see that the state with the required total angular momentum 7=1 (/ = £.-}- 5) can be realised in the following four ways: 1)1 = 0, 5=1; 2)L=1, 5 = 0; 3)Z.= 1, 5=1; 4)L = 2, 5=1. The spin functions corresponding to 5= land 5 = are respect- ively symmetric and antisymmetric in the two neutron spins. The symmetry of the coordinate functions of given L is the same as the parity (— l) L of these functions, since the reflection r t -> — r v r t -* — r t corresponds to a permutation of the coordinates of the two particles in the system of reference with its origin at r * 3* even L corre- sponds to a symmetric and odd L to an antisymmetric function of the neutron coordinates It follows thus immediately that the combinations 1, 2 and 4 are forbidden by the Pauli principle, since the coordinates and the spin functions have the same symmetry so that the total wave function is symmetric in the two neutrons. Reference to the Pauli principle shows thus that only state 3 which corresponds to an antisymmetric total wave function is poss- ible; in fact there are three such states, since 7=1. However, a tr an- SCATTERING 377 sition to such a state violates the parity conservation law since the final state is odd [(_!)* = (_ 1)1= 1]t while the initial ^- +d _ state was even, as follows from our asumptions andthe addition rule for parity. The fact that in actual fact the process considered takes place with appreciable probability is one proof that the if -meson is pseudoscalar, that is, has negative intrinsic parity. 35. * The least difference in rest mass energy of the *r and rfi meson for which the process under consideration is energetically possible is clearly equal to the sum of the deuteron binding energy (za 2.2 MeV) and the difference in rest mass energy of the neutron and the proton (* 1.25 MeV) i.e., about 3.5 MeV. (We take into account that the *" meson is captured in a bound state and possesses thus not only kinetic energy but also "collects" some energy from the reaction, though only a very sm all am ount of the order of som e ke V. ) The actual rest mass energy difference is approximately 4.5 MeV. The reaction is thus energetically possible, but the particle pro- ' duced in the reaction will possess small momenta - which is essen- tial in the following argument. We shall apply to the process considered the conservation laws of angular momentum and parity andthe Pauli principle (for neutrons). The total angular momentum / of the final system (/t-f-n-|-*<>) must be equal to the total angular momentum of the initial if~+^/-system , that is, equal to unity (see the preceding problem). Let L and 5 denote the orbital angular momentum and spin of the subsystem of two neutrons and let I denote the orbital angular momentum of the he *o -meson with respect to this subsystem - or rather with respect to its centre of mass, which is approximately the same as the cen- tre of mass of the system of all three particles. We note first of all that apart from the strict selection rules imposed by the conservation laws and the Pauli principle, there is an additional fact which strongly decreases the probability of trans- itions which satisfy the selection rules, namely, the smallness of the momenta of the neutrons and the «<> -meson - from the conser- vation of angular momentum it follows that all three momenta, P , are of the same order of magnitude. Indeed, the matrix element for any possible transition contains under the integral sign a meson -nuc- leon interaction operator, which differs appreciably from zero only in a small region of the order of the range of nuclear forces, that is, of the order of the Compton wavelength ~ of the pion. This operator acts on the wave functions of the m e son and one of the nucleon s. However , these wave functions vary in the region of interest near the origin of 37g ANSWERS AND SOLUTIONS order a~ A as fcY and (¥A so that the probability for an allowed transition contains clearly a factor (^-j It is convenient to estimate the value of (*f) as follows. In the most important case (where, as Migdal has shown, the two neu- trons fly away at a small relative angle and with a small relative kinetic energy thanks to the interaction of the two slow neutrons in the final state, so that the it°-meson flies away with a momentum which is roughly twice that of each neutron) the jf-meson carries away practically all "excess" kinetic energy A£^f 4.5 MeV — 3.5 MeV = 1 MeV ~-| where s is the deutron binding energy, since f*<C M so that p 2 ~ !*•• and hence The presence of three particles in the final state increases considerably - as compared to the preceding problem - the number of ways in which it can be realised with total angular momentum y = 1 and satisfying the Pauli principle and the parity conservation law. Since however fffj <C 1 we can henceforth consider only those allowed transitions for which the sum L + I is minimum - and the probability thus maximum. We consider two possibilities 1. The charged and neutral pions have the same intrinsic parity In this case the parity conservation law allows only even values of L + i , since the total orbital parity of the n-\-n 4-*° system, (— 1) , must be equal to +1. If we take all combinations of L, S, and I whichmake (L + /) = 0, or 2, and j = \L + S + l\ = 1. we get the following states 1)1-0. S=l, / = 0; 2)L=1, S = 0, l=U 3)L=1, S=l, /=1; 4)L=2, 5=1, / = 0; 5)1 = 0, S=l, * = 2. From these five combinations only one, namely 3), satisfies the Pauli principle (see preceding problem). In this case L + / = 2, so that the probability for the reaction considered is proportional to (£pY if the ir~ and n °-mesons have the same parity. 2. The charged and neutral mesons have opposite parity. In that case (L-f /) must be odd, since (— 1) :— 1. SCATTERING 379 If we restrict ourselves to the most probable transitions, that is, if we take L+l = 1, we get for / = 1 the following states 1)1=1, 5 = 0, / = 0; 2)Z.= 1, 5=1, / = 0; 3)L = 0, 5 = 0, /=1; 4)L = 0, 5=1, /==1. From these states 2) and 3) are allowed by the Pauli principle. Since L-\-l = X, the probability for the reaction considered is proportional to (^J if the it and «~ -meson have opposite parity; in other words, the reaction is much less forbidden than for case 1. Experiments about the capture of rc~-mesons by deuterons show that it proceeds through the reactions it" -f- d -*■ n + a or re" -f- d -*■ n -\-n -f- Tf without involving a «° -meson. This means that the reaction «"" -+- d -*» « -\~n -f- it must be strongly prohibited. From this we conclude that case 1 holds, i.e. , that the charged and the neutral pions have the same parity. It may be added that all experimental facts point to an intrinsic parity -1 for the pions. 36. * We shall apply to the reaction p-\-p-+p-{-p-\-vP (!) the laws of conservation of total angular momentum and parity and also the Pauli principle for the protons. The fact that the it°-meson is pseudoscalar means by defini- tion that its spin is equal to zero, and its intrinsic parity -1. The final state of the reaction (1) is in the case considered by us a state of orbital angular momentum L = 1 and thus of orbital parity (— 1) L = = - 1. The total parity of the final state is thus (— 1) • (— 1) = -f- 1. The parity conservation law chooses from the wave functions of the relative motion of the protons in the initial state only the spher- ical harmonics with even L (thus of parity (— l) x — -|-l) because of the one-to-one connection between orbital parity and orbital angular momentum L. As we have said several times, the orbital parity of a system of two identical particles is the same as the symmetry of their wave function with respect to the coordinates: in other words, the initial wave function of the p+p -system must be symmetric in the coord- inates and thus antisymmetric in the proton spins, because of the Pauli principle. This wave function must thus correspond to a total spin 5 = 0, and the total angular momentum, (J=S-\-L) must thus 380 ANSWERS AND SOLUTIONS be even. On the other hand, the Pauli principle admits in the final state only a state of the protons which is antisymmetric in the spins 5 = 0, because we have assumed that L = . From the rule of addi- tion of angular momenta we get for the final state the odd value 7 = 1 which violates the conservation law for total angular momentum. This concludes our proof. 37. * The nucleon isotopic spin x= ^. Hence, in complete ana- logy with the rule for adding angular momenta, the isotopic spin T of a system of two nucleons must be either unity or zero. The first of these eigenvalues is three -fold degenerate with possible values + 1,0, and -1 for r 3 : each of the three T, T 3 pairs corresponds to an isotopic function symmetric in the two nucleons. The eigen- value T = corresponds to one - antisymmetric - isotopic function. We note that the whole problem is completely analogous to the problem of combining two spins s = \. We can thus write down a table, which is immediately clear without further explanations, of the normalised isotopic functions and the corresponding values of T and T s , where we have indicated the physical meaning of each state, and where 1 and 2 denote the isotopic variable x 3 of the two nucleons. T T 9 Isotopic spin function Symmetry Interpre- tation 1 + 1 *l = *'.;: (i)^(2) symmetric 2 protons + ^(2)^ v ,(l)] it proton + neutron — 1 ^=^,(1)^,(2) •• 2 neutrons w2-^H$<i>4£*<2>- -^(2)^0)1 anti- symmetric proton + neutron From the table it is clear, in particular, that the deuteron (whose coordinate -spin state is ss-f- 8 I>)has T = 0. Indeed, its wave function corresponds to the values L = and L = 2, i. e. , it is even; its spin is 1 and it is thus symmetric also in the spins of the two nucleons. The generalised Pauli principle requires, since the total wave function is a product, that it is antisymmetric in the isotopic SCATTERING 381 variables of the nucleons; this requirement is satisfied only for 7=0. One can say that the isotopic function of the deuteron is an "isotopic scalar" that is, simply a number, as far as its transfor- mation properties are concerned, while the isotopic function of a nucleon is a two -component vector - that is, a spinor of the first rank - and the function of a pion, for instance, an ordinary vector with three components. 38. * Since the deuteron is present both in the initial and in the final state of both reactions and since for the deuteron r= T s = 0, it does not enter into the addition of the isotopic spins. We can thus simply forget about it on both sides of the reactions. In other words, the deuteron plays the role of "catalyst" for the "dissociation" process of a proton p into a nucleon and a pion, /)-> nucleon (r=-^)+ pion (7=1). (1) The left hand side of the reaction (1) has T=T 9 = ±. Since the total isotopic spin and its third component (that is, charge) are conserved, the right hand side must have the same values of Tand T z » Expanding the isotopic wave function «p$ in terms of the iso- topic functions of the "subsystems", nucleon and pion, we have The squares of the coefficients of this superposition (i and |-) give us clearly the required relative probability for the "dissocia- tion" processes P -*■ p + rc and p -> « -|- «+, and thus also the required ratio of the cross-sections. 39. * Since for the deuteron T = T 3 = 0, the left hand side of the reaction (1) corresponds to T = 1, T 9 = 1 and the left hand side of the reaction (2) to T= 1, T 3 = 0. The charge conservation lawi(r 3 = const) is clearly satisfied iden- tically. The final states of both reactions (1) and (2) are character- ised by well-defined values of the total isotopic spin T and are des- cribed respectively by the isotopic functions fj and W J. The isotopic functions of the initial states are the products of the isotopic functions of the "subsystems", that is, the nucleons and can be written (using the formulae for the addition of angular momenta) as the superposi- tion of functions of two nucleon system s with a well-defined value of T ; 382 ANSWERS AND SOLUTIONS reaction (1) : *!W£=¥i. reaction (2) n.^==-i/"i^+i/"^o- Two nucleons in the p -}-/>-state give thus exactly the state of the "isotopic triplet" 7= 1 which is necessary for reaction (1) because of the conservation laws for T and r, , while in the tt-j-p- state the triplet state T— 1 (7 , 8 = 0) which is required for reaction (2) only is present with a probability %. Since the other factors in the expression for the cross-section were assumed to be identical, the ratio of the two cross-sections is equal to the ratio of the proba- bilities of the necessary isotopic triplet state in the initial states, which is equal to two, as had to be proved. 40. * The required equality of the cross-sections expresses the charge symmetry of the nucleon-nucleon and nucleon-pion interac- tions, that is, the invariance of these interactions, the correspond- ing transition probabilities, . . . . , under a simultaneous replace- ment: n -+p, p -*■ n, % + -+ rT, tT -> « + . The equality of the cross-sec- tions follows immediately from the transformation rule of the iso- topic functions when isotopic spins are added, taking into account the generalised Pauli principle. Indeed, since the wave function of two protons or two neutrons must be antisymmetric in their spins and coordinates, it must be symmetric in their isotopic spin variables, that is, it must corre- spond to a total isotopic spin T— 1 . (This follows also very simply from the fact that T> \ T,\, where T t = + 1 for (p+p) and r,= -1 for (/! + *))• In this case the isotopic function of the p + p -system is the same as the function for * + (both are of the form W\), while the isotopic function of n-\-n is the same as that of *" (both are (W_i). The right hand sides of (1) and (2) correspond thus to the same iso- topic spin function <Fj<F _[, and the matrix elements of the transitions from the initial isotopic state $%$$ to the final state, and also the cross-sections (which are proportional to absolute square of these matrix elements) are the same. APPENDIX I A number of quantum mechanical problems are solved in the semi-classical approximation. But the semi -classical solution is valid only in regions sufficiently remote from the turning points which follow from the equation V(x) = E. Since the semi-classical solution only obtains to the left and to the right of the turning points, it is necessary to "fit" it at those points to obtain the solution in the whole of space. a Fig. 44 If the potential function V(x) near the turning point behaves as shown in figure 43 the fitted solution is of the form sin ¥1=' yy-\j\ pdx +i) if x<a ' X i| dx 1 2 VTTi' if x > a. (1) (2) For the potential function V(x) shown in figure 44 the solution will be <h = Tj-ty^+f) if *>•• (i,) a 2Y\P\ if x<a 384 APPENDIX I Using these expressions we find a solution of the one -dimensional SchrSdinger equation which to the left of the turning point (see fig. 43) goes over into the semi-classical solution of the form pdx Yp It is necessary to find another solution which is linearly independent with respect to the solutions (1), (2). We can write it in the form <h={ cos yr-\ij pdx +i) if x<a ' OB (3) \dx c r=-e if jc > a. VTF\" ir x>a - (4) To determine c we use the fact that for our SchrBdinger equation W = <}>2 "W = const. We write out the Wronskian for our case, using the solutions (1) and (3), W= pdx+ T -±=sin mpdx-hj ll-^cos ij. ^ x ' 'a? y^ C0S [ij pdx ^TJ'' nr^TJp^ + T) (It is only necessary to differentiate the argument of the trigonome- trie functions, since —^~ <C !•) Similarly we see that in the region x > a that is, for the solutions (2) and (4) the Wronskian n From the condition W(x < a) = W(x > a) we find that c = 1. We get thus for the required solution a linear combination of ^ and <V 2 : APPENDIX I 385 * = (■!>.. ± i'b { ) c f ' i We have finally Vi» T.f i,rf ' T.f^« '< 1 dv ± /!/>! 2/1/; if .v < «, .7: a if x > a. We now find a solution of the one -dimensional Schrodinger equation Ifi d'^j , , T , which to the left of the turning point (see fig. 44) goes over into the semi -classical solution of the form *-/ P | tie Y\p e x To do this we use our earlier results. The required solution is namely a linear combination of the solutions £ and < found above where ' ' ' V> dx «= p a Y~p a . r. "7 i — 4 1 ] / if x > a, a K = Y\p\ X 1 * ^ /> a Yp a ~j\p\dx - p a; if x<a, if x > a, « ^l/>| — _ e *Y\p\ if x < a. 386 We get APPENDIX I \dx if x < a, IP 4= < 2Vp 2 if a; > a, |p I da? if a; < a, y = ! tJ* *"-*!■ ■J-/** r+ *T V* =r-0 y^ if a: > a. APPENDIX II A whole set of experimental data (for instance, the scattering of pions by protons and neutrons) shows that the proton and neutron can be mutually interchanged. This is the basis of considering the proton and the neutron to be one particle - a nucleon - which can be in two states: either a proton or a neutron. These states differ by the value of the charge variable: the proton has a charge 1 (in units e) and the neutron a charge 0. The nucleon can then be described by a two-component wave function, corresponding to the two possible values of the charge variable. We write this function as follows '% JtnJ Using the norm alisation condition \%\ 2 -\-\ty n \ 2 = I, we put 0/ ^ll ^ = ln/' *» = We introduce operators, acting upon these two-component functions, <+=(o J) 1 One sees easily that From these relations it follows that t + is the charge -creation op- erator, which raises a nucleon from the neutron into the proton state and t_ is the charge-annihilation operator. We now introduce the operators ^=y('++0=2-( 1 oj' 1 . . 1/1 0\ *. = T (T+T_— T_T + )=2^ _ i y 388 APPENDIX II These operators are identical with the well-known Pauli spin matrices andthey have thus the same formal properties as those. By analogy with the spins, we shall say that ? x , - tJ , and t_ are the operators of the components of the vector x in some three -dim en - sional space. This space is called isotopic space and the vector x is called the isotopic spin vector of the nucleon. We must note that the con- cept of an isotopic space is an auxiliary one and that it has no direct physical meaning. The absolute value of the vector x is \ and the two charge states of the nucleons can be considered to be states with different values of the z -component of the isotopic spin (in isotopic space). The proton corresponds to x z -\, and the neutron to T ? = h' We note that, as the isotopic space andthe isotopic spin have a formal character, there is no direct physical meaning to be attached to the operator -,, or the so-called charge operator ? = ' c = + 2 / =(o 0> The 7r + -, 7T°-, and n~ mesons can also be considered to be one par- ticle which can occur in three charge states corresponding to the values 4 1, 0, and - 1 of the charge parameter. The wave function of the pion will clearly be a three -component one, For normalising purposes we put T+ -(i) *-(D We can introduce charge -creation and annihilation operators, f° 1 Jo 1 \o which satisfy the relations 7 + cp + = 0; T +?0 = cp + ; T + <?_ = <p ; T_y + = < ?0 ; r..cp = cp_; T_y_ = 0. APPENDIX II 389 We can construct the operators /0 1 0\ ,0 1 0. r,-^(r + -r_,_^j which are the operators of the components of the isotopic spin vector T in isotopic space. The absolute magnitude of T is 1. Different charge states will be states with different values of the z -compon- ent of the isotopic spin. We note that now the charge operator is the same as T s , Let us now consider the isotopic properties of the nucleon-pion system. This system can be characterised by a total isotopic spin and its z -component I s . Present experimental evidence suggests that for the pion- nucleon systems the hypothesis of isotopic invariance (charge in- dependence) is valid, that is, the hypothesis that the properties of the system are independent of the total charge of the system, provided Coulomb forces are unimportant. Mathematically one can express this as the invariance of the interaction Hamiltonian under a rotation in isotopic space. It follows then immediately that the total isotopic spin / and its z -component I g are conserved in the pion-nucleon system. INDEX The numbers refer to the problems; for instance, 5.14 is problem 14, Section 5. Adiabatic perturbations, 3.45-48. Angular momentum, Section 4. Alpha decay, 5.14. Asymmetric top, 8.20-22. Beta decay, 7.12. Born approximation, 9.5, 9.8, 9.9, 9.28, 9.29. Commutation relations, Section 3. Commutators, 3.23, 3.24. Detailed balancing, 9.32. Deuteron, 4.32-34, 5.9, 5.10, 9.32, 9.34, 9.35, 9.38, 9.39. Diatomic molecules, 8.1-12, 8.23, 8.24, 8.26, 8.27, 8.34-36. Dipole moment, 7.67-69. Dirac equation, 7.10. Exchange effects, 7.81. Finite mass of nucleus, 7.80, 7.81, Finite size of nucleus, 7.26. Gravitational field, 1.24, 1.27. Green function, 3.10, 3.14-16, 3.44. 392 INDEX Harmonic oscillator, 1.5-7, 1.9, 1.18, 1.20, 3.3, 3.10, 3.12-17, 3,20, 3.22, 3.42, 3.44, 3.47, 5.5-7, 6.6, 7.72, 7.73. Heisenberg relations, Section 3. Heisenberg representation, 3.41-44. Helium atom, 7.14, 7.15, 7.17, 7.33, 9.9. Helium -like atoms, 7.13, 7.71, 7.81. Hermitean operators, 3.26. Hund rules, 7.37. Hydrogen atom, 3.29, 7.3-11, 7.28-30, 7.53, 7.56-59, 7.61-64, 7.66, 7.74, 7.77, 9.9. Hydrogen-like atoms, 7.1, 7.2, 7.49, 7.51, 7.52. Hydrogen molecule, 9.14. Hyperfine structure, 7.30, 7.31, 7.54. Integrals of motion, 3.30, 3.31. Internal conversion, 7.75. Ionisation potential, 7.71. Isotope shift, 7.80, 7.81. Isotopic spin, 9.21, 9.22, 9.37, App.II. Kronig-Penney potential, 1.16. Lande factor, 7.47, 7.49, 7.50. Lithium atom, 7.16. London - van der "Waals 1 forces, 8.34, 8.35. Magnetic field, motion in, Section 6. Magnetic moment, 4.30-32. Matrices, 3.35, 3.36. Mesons, 7.65, 7.66. Meson scattering, 9.22-27, 9.33-36. Momentum distribution function, 1.3, 1.9, 3.7, 3.8, 3.32-34, 7.5. Momentum representation, 1.9, 1.14, 1.15, 3.34, 3.35. Neutron quadrupole moment, 4.34. INDEX 393 Parity, 7.67, 9.35-40. Periodic potential, 1.14-16, 1.28. Perturbation theory, 5.16, 7.44-46, 7.53, 7.54, 7.60, 7.69-74, 9.31. Plane rotator, 1.23, 3.38, 7.69. Polarisability, 7.61, 7.79. Quadrupole moment, 4.34, 4.35, 7.11. Reflection coefficient, 2.1, 2.2, 2.4, 2.13, 6.16. Rotations, 4.44, 4.46. Scattering, Section 9. Scattering length, 9.16. Selection rules, 5.16, 9.33. Semi -classical approximation, 1.17-22, 1.27-29, 2.6, 2.9, 2.11, 2.12, 2.14, 2.15, 5.2, 6.10, 6.11, 7.27, 7.78, App.I. Shell model of the nucleus, 5.7, 5.8. Spherical well, 5.11-13. Spin, Section 4. Spin-orbit coupling, 7.10, 7.45. Square-well potential, 1.1-3, 1.25, 1.26, 3.32, 3.35, 3.39, 3.48. Stark effect, 7.57-59, 7.62, 7.69, 7.70, 7.72, 7.73, 8.31, 8.32. Stern -Gerlach experiment, 4.8, 4.12, 4.18. Sum rule, 7.68. Symmetric top, 8.16-19. Term schemes, 7.35-48. Thomas-Fermi atom, 7.19-25, 7.78, 7.79. Time -dependence of operators, 3.19, 3.20, 3.25, 3.38, 3.39, 3.43 3.44. Transition probabilities, 3.12, 3.17. Transmission coefficient, 2.3, 2.5-7, 2.12-15. Tunnel effect, Section 2, 5.14. Uniform field, 1.13. 394 INDEX Variational method, 5.10, 7.13, 7.16, 7.21, 7.22, 7.60, 7.61. Virialtheroem, 1.21, 3.37, 7.18, 7.23, 7.24. Wave packets, 3.9, 3.10. WKB-method, see Semi -classical approximation. Yukawa potential, 9.8, 9.28. Zeeman effect, 7.54, 7.62, 8.27-30. This collection of problems (with solutions) may be used in con- junction with Landau and Lifshits* "Quantum Mechanics" or any other textbook such as the ones by Kramers and Schiff. It may also be used independently, for advanced reading, by students familiar with the basic ideas of quantum mechanics. * The 333 problems are arranged in nine chapters and are of vary- ing difficulty; some involve simple derivations or proofs while for others the most advanced methods, both mathematical and physical, are needed. The emphasis is throughout on the building up of an under- standing of the basic ideas of quantum mechanics and on the familiari- sation with the basic techniques. In Chapter 1, solutions of the one -dimensional Schroedinger - , equation are discussed both in coordinate and in momentum represen- ,^r tations. In this chapter, as in others, many problems involve the use of the WKB or semi -classical approximation. Chapter 2 is devoted to barrier -penetration effects. Chapter 3 contains problems on commutation relations, the spreading of wave packets, the use of Green functions, operators in the Heisenberg representation, and the relation between invariance properties and constants of motion. Chapter 4 deals with angular momentum and spin; there are also some problems on quadrupole moments and on the magnetic moments of nuclei. Chapter 5 relates to the motion of a particle in a central field of force. Chapter 6 deals with the motion of particles in a magnetic field, the change of the commutation relations when a magnetic field is pre- sent, and the influence of the magnetic field on such properties as the spreading of wave packets, the reflexion of neutrons, the transitions from one spin state to another, and the orbitals of particles. Chapter 7, the longest one in the volume, deals with atomic pro- r^rties, such as the charge density in an atom, spin-orbital coupling, variational methods, the Thomas-Fermi model, spectroscopic term schemes, parity of states, Zee man and Stark effect in weak, inter- mediate, and strong fields, sum rules, and perturbation theory. Chapter 8 is devoted to molecules. Chapter 9 contains problems on scattering and isotopic spin. There are two Appendices dealing with the semi -classical approxi- mation and with isotopic spin, respectively. \ infosearch